{"id":219708,"date":"2025-05-26T15:24:56","date_gmt":"2025-05-26T15:24:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219708"},"modified":"2025-05-26T15:24:58","modified_gmt":"2025-05-26T15:24:58","slug":"ammonium-phosphate-nh43po4-is-an-important-ingredient-in-many-fertilizers","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/26\/ammonium-phosphate-nh43po4-is-an-important-ingredient-in-many-fertilizers\/","title":{"rendered":"Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers"},"content":{"rendered":"\n

Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with ammonia NH3 . What mass of phosphoric acid is consumed by the reaction of 4.10g of ammonia?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the mass of phosphoric acid (H\u2083PO\u2084<\/strong>) consumed in the reaction with 4.10 g of ammonia (NH\u2083<\/strong>), we start by writing the balanced chemical equation:<\/p>\n\n\n\n

H\u2083PO\u2084 + 3NH\u2083 \u2192 (NH\u2084)\u2083PO\u2084<\/strong><\/p>\n\n\n\n

Step-by-Step Calculation:<\/h3>\n\n\n\n
    \n
  1. Determine the molar mass of NH\u2083<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • N: 14.01 g\/mol<\/li>\n\n\n\n
    • H: 1.008 g\/mol \u00d7 3 = 3.024 g\/mol<\/li>\n\n\n\n
    • Molar mass of NH\u2083 = 14.01 + 3.024 = 17.034 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
        \n
      1. Calculate moles of NH\u2083<\/strong>:<\/li>\n<\/ol>\n\n\n\n

        Moles of NH\u2083=4.10 g17.034 g\/mol\u22480.2407 mol\\text{Moles of NH\u2083} = \\frac{4.10\\ \\text{g}}{17.034\\ \\text{g\/mol}} \\approx 0.2407\\ \\text{mol}<\/p>\n\n\n\n

          \n
        1. Use the molar ratio from the balanced equation<\/strong>:
          From the reaction:<\/li>\n<\/ol>\n\n\n\n
            \n
          • 1 mol of H\u2083PO\u2084 reacts with 3 mol of NH\u2083
            So, the number of moles of H\u2083PO\u2084 needed:<\/li>\n<\/ul>\n\n\n\n

            Moles of H\u2083PO\u2084=0.2407 mol NH\u20833\u22480.0802 mol\\text{Moles of H\u2083PO\u2084} = \\frac{0.2407\\ \\text{mol NH\u2083}}{3} \\approx 0.0802\\ \\text{mol}<\/p>\n\n\n\n

              \n
            1. Calculate the mass of H\u2083PO\u2084<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                \n
              • H\u2083PO\u2084 molar mass:\n
                  \n
                • H: 1.008 \u00d7 3 = 3.024<\/li>\n\n\n\n
                • P: 30.97<\/li>\n\n\n\n
                • O: 16.00 \u00d7 4 = 64.00<\/li>\n\n\n\n
                • Total = 3.024 + 30.97 + 64.00 = 97.994 g\/mol<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

                  Mass of H\u2083PO\u2084=0.0802 mol\u00d797.994 g\/mol\u22487.86 g\\text{Mass of H\u2083PO\u2084} = 0.0802\\ \\text{mol} \u00d7 97.994\\ \\text{g\/mol} \\approx 7.86\\ \\text{g}<\/p>\n\n\n\n

                  \u2705 Final Answer:<\/h3>\n\n\n\n

                  7.86 grams of phosphoric acid (H\u2083PO\u2084)<\/strong> are consumed.<\/p>\n\n\n\n

                  \n\n\n\n

                  Explanation:<\/h3>\n\n\n\n

                  The reaction between phosphoric acid (H\u2083PO\u2084) and ammonia (NH\u2083) forms ammonium phosphate ((NH\u2084)\u2083PO\u2084), a key component in many fertilizers. This process is important in agriculture, where fertilizers are used to supply plants with essential nutrients like nitrogen and phosphorus. Understanding the stoichiometry of this reaction allows us to calculate how much of one reactant is needed to completely react with a given amount of another.<\/p>\n\n\n\n

                  The chemical reaction is:
                  H\u2083PO\u2084 + 3NH\u2083 \u2192 (NH\u2084)\u2083PO\u2084<\/strong><\/p>\n\n\n\n

                  This balanced equation tells us that one mole of phosphoric acid reacts with three moles of ammonia. We begin by converting the given mass of ammonia (4.10 grams) into moles using its molar mass, which is 17.034 g\/mol. This gives approximately 0.2407 moles of NH\u2083.<\/p>\n\n\n\n

                  Next, we apply the mole ratio from the balanced equation. Since it takes three moles of ammonia to react with one mole of phosphoric acid, we divide the ammonia moles by three. This gives about 0.0802 moles of H\u2083PO\u2084 required.<\/p>\n\n\n\n

                  Then, we convert these moles into grams using the molar mass of H\u2083PO\u2084 (97.994 g\/mol). Multiplying gives approximately 7.86 grams of H\u2083PO\u2084.<\/p>\n\n\n\n

                  This method demonstrates the core principle of stoichiometry: using balanced equations and molar relationships to predict the outcomes of chemical reactions. This type of calculation is critical in industrial applications like fertilizer production, where precise amounts of reactants must be mixed to maximize yield and minimize waste.<\/p>\n\n\n\n

                  \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                  Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with ammonia NH3 . What mass of phosphoric acid is consumed by the reaction of 4.10g of ammonia? The Correct Answer and Explanation is: To find the mass of phosphoric acid (H\u2083PO\u2084) consumed in the reaction […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-219708","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219708","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=219708"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219708\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219708"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219708"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}