{"id":7986,"date":"2013-09-07T00:22:17","date_gmt":"2013-09-07T00:22:17","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=7986"},"modified":"2026-03-26T12:53:32","modified_gmt":"2026-03-26T17:53:32","slug":"concept-of-a-mole-ratio","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/concept-of-a-mole-ratio\/","title":{"rendered":"Mole Ratio"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_Mg9aPUrf6gE\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_Mg9aPUrf6gE\" data-source-videoID=\"Mg9aPUrf6gE\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Mole Ratio Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Mole Ratio\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_Mg9aPUrf6gE:hover {cursor:pointer;} img#videoThumbnailImage_Mg9aPUrf6gE {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1441-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_Mg9aPUrf6gE\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_Mg9aPUrf6gE\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Mole Ratio\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_Mg9aPUrf6gE\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_Mg9aPUrf6gE\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_Mg9aPUrf6gE\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction fCo_Function() {\n  var x = document.getElementById(\"fCo\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"fCo_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"fCo\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example #1<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example #2<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_3\" class=\"smooth-scroll\">Example #3<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_4\" class=\"smooth-scroll\">Example #4<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_5\" class=\"smooth-scroll\">Example #5<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video on the mole ratio, sometimes referred to as the molar ratio.<\/p>\n<p>It\u2019s a fundamental concept in chemistry that\u2019s used to convert back and forth between moles of each substance in a reaction. Among other things, it\u2019s necessary for calculating theoretical yields and limiting reagents.<\/p>\n<p>This is most certainly a concept best learned through examples, so let\u2019s jump right in.<\/p>\n<h2><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h2>\n<p>\nBefore we put these mole ratios to use, let\u2019s consider a more relatable example to solidify this concept.<\/p>\n<p>When you make mac n\u2019 cheese, you always use the same ratio of ingredients. One box of pasta, 1 packet of cheese, 3 tablespoons of milk, and 2 tablespoons of butter.<\/p>\n<div class=\"transcriptcallout\">1 box pasta + 1 packet of cheese + 3 tbsp milk + 2 tbsp butter = 1 mac n&#8217;cheese<\/div>\n<p>\n&nbsp;<br \/>\nOur \u201cmole ratio\u201d, so to speak, for milk and mac n\u2019 cheese is 1 mac n\u2019 cheese over 3 tablespoons of milk.<\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{1\\text{ mac n&#8217;cheese}}{3\\text{ tbsp milk}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, for every 3 tablespoons of milk, we can make one recipe of mac n\u2019 cheese. <\/p>\n<p>We could do this for any combination of species in the reaction. For example, for every packet of cheese we have, we need 2 tablespoons of butter. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{2\\text{ tbs butter}}{1\\text{ packet of cheese}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd of course, if we want more mac n\u2019 cheese we can double the recipe.<\/p>\n<div class=\"transcriptcallout\">2 boxes pasta + 2 packets of cheese + 6 tbsp milk + 4 tbsp butter = 2 mac n&#8217;cheese<\/div>\n<p>\n&nbsp;<br \/>\nIf we did that, the ratio between ingredients would stay the same. <\/p>\n<p>So, in fact, anytime you follow a recipe, you\u2019re basically employing the concept of mole ratios, so keep that in mind the next time you\u2019re cooking!<\/p>\n<h2><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h2>\n<p>\nNow let\u2019s put these mole ratios to work by looking at the formation of ammonia from nitrogen and hydrogen. <\/p>\n<div class=\"examplesentence\">\\(\\text{N}_2 + \\text{H}_2 \u21cc \\text{NH}_3\\)<\/div>\n<p>\n&nbsp;<br \/>\nFirst, as written, this is an unbalanced chemical equation. On the reactants side of the equation, we have 2 atoms of nitrogen and 2 atoms of hydrogen, while on the product side, we only have 1 atom of nitrogen but 3 atoms of hydrogen. <\/p>\n<p>We can\u2019t even discuss the concept of the mole ratio for a reaction if it is unbalanced. So let\u2019s balance this first.<\/p>\n<p>Remember, we can\u2019t change the molecules, split bonds, or change the chemistry in any way. We\u2019re stuck working with the molecules we\u2019re given, so to balance a reaction we can only increase the number of each molecule by adjusting the stoichiometric coefficients. <\/p>\n<p>To balance this reaction, we need 1 molecule of nitrogen gas, 3 molecules of hydrogen gas, and 2 molecules of ammonia. Now we have 2 nitrogen atoms and 6 hydrogen atoms on both sides of the equation. <\/p>\n<div class=\"examplesentence\">\\(\\text{N}_2 + 3\\text{H}_2 \u21cc 2\\text{NH}_3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSaid another way, to make 2 molecules of ammonia, we need 1 molecule of nitrogen and 3 molecules of hydrogen. To have a balanced reaction with no leftover atoms, the chemistry demands this ratio of molecules. <\/p>\n<p>We could, of course, multiply by a factor, say of 2, so that 2 molecules of nitrogen plus 6 molecules of hydrogen make 4 molecules of ammonia, which still works because we haven\u2019t changed the underlying ratio between molecules. In other words, 1 divided by 3 equals 2 divided by 6. <\/p>\n<p>We can extend this to the mole scale, as in, when we think about zillions of molecules reacting at once. Remember, a mole is an amount of something equivalent to 6.022 \u00d7 10<sup>23<\/sup> units, which could be atoms, ions, electrons, or, like in this case, molecules.<\/p>\n<p>So when we say that 1 mole of nitrogen is needed to react with 3 moles of hydrogen to make 2 moles of ammonia, we\u2019re saying that we need 6.022 \u00d7 10<sup>23<\/sup> molecules of nitrogen to react with 18.066 \u00d7 10<sup>23<\/sup> molecules of hydrogen to make 12.044 \u00d7 10<sup>23<\/sup> molecules of ammonia. While those are much larger numbers than 1, 2, and 3, the ratio is maintained between the molecules. <\/p>\n<p>Another way of phrasing this is that 1 mole of nitrogen is stoichiometrically equivalent to 3 moles of hydrogen when making ammonia. This means that 1 mole of nitrogen gas will make you the same amount of ammonia as 3 moles of hydrogen. So even though we have different amounts of each substance, when it comes to making ammonia, we can think of them as being equivalent. <\/p>\n<p>We can actually write this out in an equation. <\/p>\n<p>For the formation of ammonia, one mole of nitrogen gas is stoichiometrically equivalent to 3 moles of hydrogen gas.<\/p>\n<div class=\"examplesentence\">\\(1\\text{ mol N}_2 = 3 \\text{ mol H}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can then rearrange this to get the actual mole ratio 1 mole of nitrogen gas over 3 moles of hydrogen gas is equal to 1.<\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{1\\text{ mol N}_2}{3\\text{ mol H}_2}=1\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe now can use this to convert between moles of hydrogen to moles of nitrogen, which will be very useful when we need to calculate yields or limiting reagents. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{3\\text{ mol H}_2}{1\\text{ mol N}_2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOf course, we could flip the ratio if we needed to convert from moles of nitrogen to moles of hydrogen. <\/p>\n<p>Similarly, we can write out the molar ratios for the other two combinations of reactants and products. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{1\\text{ mol N}_2}{2\\text{ mol NH}_3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAs you can see here, for every 1 mole of nitrogen, we can make 2 moles of ammonia <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{3\\text{ mol H}_2}{2\\text{ mol NH}_3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nand for every 3 moles of hydrogen, we can make 2 moles of ammonia.<\/p>\n<h2><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h2>\n<p>\nNow let\u2019s think about how we actually use mole ratios in chemistry problems.<\/p>\n<p>Let\u2019s say we wanted to make 10 moles of ammonia- how many moles of nitrogen and hydrogen would we need? With our mole ratios, this is a simple problem. Let\u2019s work out the hydrogen first. 10 moles of ammonia times 3 moles of hydrogen gas over 2 moles of ammonia equals 15 moles of hydrogen gas.<\/p>\n<div class=\"examplesentence\">\\(10\\text{ mol NH}_3(\\frac{3\\text{ mol H}_2}{2\\text{ mol NH}_3})=15\\text{ mol H}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nOkay, now let\u2019s work out the nitrogen. 10 moles of ammonia times 1 mole of nitrogen gas over 2 moles of ammonia equals 5 moles of nitrogen gas.<\/p>\n<div class=\"examplesentence\">\\(10\\text{ mol NH}_3(\\frac{1\\text{ mol N}_2}{2\\text{ mol NH}_3})=5\\text{ mol N}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, to make 10 moles of ammonia, we need 15 moles of hydrogen and 5 moles of nitrogen.<\/p>\n<p>Notice that when you write these calculations out, it\u2019s very useful to specify the chemical species to which you are referring. So rather than just writing \u201c4 moles\u201d, write \u201c4 moles NH<sub>3<\/sub>\u201d to keep track of each.<\/p>\n<h2><span id=\"Example_4\" class=\"m-toc-anchor\"><\/span>Example #4<\/h2>\n<p>\nLet\u2019s move on to a slightly more complicated problem. Say we had 1 kilogram of nitrogen gas and wanted to know how many kilograms of ammonia we could make with that (assuming we have unlimited amounts of hydrogen). <\/p>\n<p>To calculate this, we first need to know how many moles of nitrogen gas are in 1 kg. For that conversion, we use the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/calculating-the-molar-mass-of-a-substance\/\">molar mass<\/a> of nitrogen gas, which is 28.0 grams per mole. So, 1kg of nitrogen times 1,000 grams per 1 kg times 1 mole of nitrogen per 28 grams of nitrogen equals 35.7 moles of nitrogen.<\/p>\n<div class=\"examplesentence\">\\(1.00\\text{ kg N}_2(\\frac{1000 \\text{g}}{1\\text{ kg}})(\\frac{1\\text{ mol N}_2}{28.0\\text{ g N}_2})=35.7\\text{ moles N}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we have the moles of nitrogen, we must convert to moles of ammonia. And here we use our mole ratio. 35.7 moles of nitrogen times 2 moles of ammonia over 1 mole of nitrogen equals 71.4 moles of ammonia.<\/p>\n<div class=\"examplesentence\">\\(35.7\\text{ moles N}_2(\\frac{2\\text{ mol NH}_3}{1\\text{ mol N}_2})=71.4\\text{ moles NH}_3\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd lastly, we convert the moles of ammonia to kilograms of ammonia with its molar mass, which is 17.0 grams per mole. So, 71.4 moles of ammonia times 17 grams of ammonia over 1 mole of ammonia times 1 kg over 1,000 grams equals 1.21 kg of ammonia.<\/p>\n<div class=\"examplesentence\">\\(71.4\\text{ moles NH}_3(\\frac{17.0\\text{ g NH}_3}{1\\text{ mol NH}_3})(\\frac{1\\text{ kg}}{1000\\text{ g}})=1.21\\text{ kg NH}_3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, using the mole ratio, we were able to calculate that 1 kg of nitrogen can make 1.21 kg of ammonia.<\/p>\n<p>Now, pause the video and take a minute to calculate how many grams of hydrogen would be needed to make 500 grams of ammonia, assuming that nitrogen is in excess. The molar masses of ammonia and hydrogen are 17 g\/mol and 2 g\/mol, respectively.<\/p>\n<p>Hopefully, that felt pretty straightforward (if not- don\u2019t worry, we have some more practice coming soon). Here&#8217;s how we solved that problem. <\/p>\n<div class=\"examplesentence\">\\(500\\text{ g NH}_3(\\frac{1\\text{ mol NH}_3}{17.0\\text{ g NH}_3})(\\frac{3\\text{ mol H}_2}{2\\text{ mol NH}_3})(\\frac{2.00\\text{ g H}_2}{1\\text{ mol H}_2})=88.2\\text{ g H}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nBegin with 500 grams of ammonia, convert to moles with the molar mass, then convert to moles of hydrogen with the mole ratio, and finally, convert to grams of hydrogen with the molar mass of hydrogen. That gives us 88.2 grams of hydrogen needed to make 500 grams of ammonia. <\/p>\n<p>Alright, before we look at a practice problem, let\u2019s quickly review what we\u2019ve learned.<\/p>\n<p>Mole ratios are used to convert between moles of different species in a given chemical reaction. Always make sure that your chemical reaction is balanced because mole ratios are derived from the stoichiometric coefficients. So if the reaction isn\u2019t balanced, you won\u2019t get the correct mole ratios!<\/p>\n<h2><span id=\"Example_5\" class=\"m-toc-anchor\"><\/span>Example #5<\/h2>\n<p>\nNow try your hand at this problem:<\/p>\n<p>For the combustion of propane, answer the following questions<\/p>\n<div class=\"examplesentence\">\\(\\text{C}_3\\text{H}_8+5\\text{O}_2\\rightarrow 3\\text{CO}_2+4\\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\n1. What is the mole ratio between oxygen and water?<br \/>\n2. If you burned 10 moles of propane, how many moles of carbon dioxide would you emit into the atmosphere?<br \/>\n3. If you wanted to combust 1 lb of propane, how many moles of oxygen would you need? <\/p>\n<p>(1.00 lb = 454 grams, and the molar mass of propane = 44.1 g\/mol)<\/p>\n<p>Go ahead and pause the video now if you want to try this one on your own.<\/p>\n<p>Alright, here\u2019s how you should approach this. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-71821\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/mole-ratio-between-oxygen-and-water.png\" alt=\"mole ratio between oxygen and water\" width=\"527\" height=\"221\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/mole-ratio-between-oxygen-and-water.png 890w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/mole-ratio-between-oxygen-and-water-300x126.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/mole-ratio-between-oxygen-and-water-768x322.png 768w\" sizes=\"auto, (max-width: 527px) 100vw, 527px\" \/><\/p>\n<p>First, check to make sure that the reaction was balanced- which it was. <\/p>\n<p>1. Then, to find the mole ratio between oxygen and water, simply use the stoichiometric coefficients to show that for every 5 moles of oxygen that you combust, you produce 4 moles of water. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{5\\text{ mol O}_2}{4\\text{ mol H}_2\\text{O}}\\) or \\(\\frac{4\\text{ mol H}_2\\text{O}}{5\\text{ mol O}_2}\\)<\/div>\n<p>\n&nbsp;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-71827\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/moles-of-oxygen-and-water.png\" alt=\"moles of oxygen and water\" width=\"208\" height=\"178\" \/><\/p>\n<p>2. To convert moles of propane to moles of carbon dioxide, you need to use their mole ratio. In this case, for every mole of propane, you produce 3 moles of carbon dioxide. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{3\\text{ mol CO}_2}{1\\text{ mol C}_3\\text{H}_8}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince you started with 10 moles of propane, this means you would produce 30 moles of carbon dioxide. <\/p>\n<div class=\"examplesentence\">\\(10\\text{ mol C}_3\\text{H}_8(\\frac{3\\text{ mol CO}_2}{1\\text{ mol C}_3\\text{H}_8})=30\\text{ mol CO}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-71830\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/moles-of-propane-and-carbon-dioxide.png\" alt=\"10 moles of propane and 30 moles carbon dioxide\" width=\"286\" height=\"111\" \/><\/p>\n<p>3. And lastly, to answer the third question, we convert pounds of propane to grams of propane, then to moles of propane. We then use the mole ratio to convert to moles of oxygen needed.<\/p>\n<div class=\"examplesentence\">\\(1.00\\text{ lb C}_3\\text{H}_8(\\frac{454\\text{ g}}{1\\text{ lb}})(\\frac{1\\text{ mol C}_3\\text{H}_8}{44.1\\text{ g C}_3\\text{H}_8})(\\frac{5\\text{ mol O}_2}{1\\text{ mol C}_3\\text{H}_8})=51.5\\text{ mol O}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis shows us that we would need 51.5 moles of oxygen to react with 1 lb of propane. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-71824\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/04\/moles-of-oxygen-and-propane.png\" alt=\"5 moles of oxygen and 4 moles of propane\" width=\"281\" height=\"168\" \/><\/p>\n<p>Okay, that\u2019s all for our review of mole ratios! Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/chemistry\/\">Return to Chemistry Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Chemistry Videos<\/p>\n","protected":false},"author":1,"featured_media":100486,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-7986","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-atoms-ions-and-molecules-videos","7":"page_category-chemistry-instruments-and-calculations","8":"page_type-video","9":"subject_matter-science"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/7986","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=7986"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/7986\/revisions"}],"predecessor-version":[{"id":280340,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/7986\/revisions\/280340"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100486"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=7986"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}