{"id":192513,"date":"2025-02-18T04:01:59","date_gmt":"2025-02-18T04:01:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192513"},"modified":"2025-02-18T04:02:01","modified_gmt":"2025-02-18T04:02:01","slug":"ammonium-phosphate-fertilizer-is-prepared-by-the-following-reaction","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/18\/ammonium-phosphate-fertilizer-is-prepared-by-the-following-reaction\/","title":{"rendered":"Ammonium phosphate fertilizer is prepared by the following reaction"},"content":{"rendered":"\n

Ammonium phosphate fertilizer is prepared by the following reaction: 3NH3(g) + H3PO4(aq) \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae (NH4)3PO4(s) (a) \u00c3\u201a\u00c2 Calculate the molar mass of ammonia, phosphoric acid and ammonium phosphate. (b) \u00c3\u201a\u00c2 How many moles of ammonium phosphate are present in a 5.00-pounds bag of (NH4)3PO4?
Ammonium phosphate fertilizer is prepared by the following reaction: 3NH3(g) + H3PO4(aq) \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae (NH4)3PO4(s)<\/p>\n\n\n\n

(a) \u00c3\u201a\u00c2 Calculate the molar mass of ammonia, phosphoric acid and ammonium phosphate.<\/p>\n\n\n\n

(b) \u00c3\u201a\u00c2 How many moles of ammonium phosphate are present in a 5.00-pounds bag of (NH4)3PO4?<\/p>\n\n\n\n

(c) \u00c3\u201a\u00c2 How many grams of ammonia and phosphoric acid, respectively, are required to produce 1.00 kg of<\/p>\n\n\n\n

(NH4)3PO4 if the reaction has 100% yield;<\/p>\n\n\n\n

(d) \u00c3\u201a\u00c2 How many kilograms (kg) of ammonia and phosphoric acid, respectively, are required to produce<\/p>\n\n\n\n

1.00 kg of (NH4)3PO4 if the reaction has has 85.0% yield?<\/p>\n\n\n\n

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

Let’s solve each part step by step:<\/p>\n\n\n\n

(a) Calculate the molar mass of ammonia, phosphoric acid, and ammonium phosphate.<\/h3>\n\n\n\n
    \n
  1. Ammonia (NH\u2083):<\/strong>
    • Nitrogen (N) has an atomic mass of 14.01 g\/mol.<\/li>
    • Hydrogen (H) has an atomic mass of 1.008 g\/mol.<\/li><\/ul>Molar mass of NH\u2083 = 14.01 + (3 \u00d7 1.008) = 17.034 g\/mol<\/strong>.<\/li>\n\n\n\n
    • Phosphoric acid (H\u2083PO\u2084):<\/strong>
      • Hydrogen (H) has an atomic mass of 1.008 g\/mol.<\/li>
      • Phosphorus (P) has an atomic mass of 30.974 g\/mol.<\/li>
      • Oxygen (O) has an atomic mass of 16.00 g\/mol.<\/li><\/ul>Molar mass of H\u2083PO\u2084 = (3 \u00d7 1.008) + 30.974 + (4 \u00d7 16.00) = 97.994 g\/mol<\/strong>.<\/li>\n\n\n\n
      • Ammonium phosphate ((NH\u2084)\u2083PO\u2084):<\/strong>
        • From NH\u2084 (ammonium) and PO\u2084 (phosphate), we know the molar mass is calculated by adding up the masses of 3 ammonium ions and 1 phosphate ion.<\/li><\/ul>Molar mass of (NH\u2084)\u2083PO\u2084 = 3 \u00d7 (14.01 + 4 \u00d7 1.008) + 30.974 + 4 \u00d7 16.00 = 3 \u00d7 18.034 + 30.974 + 64.00 = 149.086 g\/mol<\/strong>.<\/li>\n<\/ol>\n\n\n\n

          (b) How many moles of ammonium phosphate are present in a 5.00-pound bag of (NH\u2084)\u2083PO\u2084?<\/h3>\n\n\n\n

          To convert 5.00 pounds to grams: 1\u2009lb=453.592\u2009g1 \\, \\text{lb} = 453.592 \\, \\text{g} 5.00\u2009lb=5.00\u00d7453.592=2267.96\u2009g5.00 \\, \\text{lb} = 5.00 \\times 453.592 = 2267.96 \\, \\text{g}<\/p>\n\n\n\n

          Now, use the molar mass of ammonium phosphate (149.086 g\/mol) to find moles: Moles of (NH4)3PO4=2267.96\u2009g149.086\u2009g\/mol=15.21\u2009mol\\text{Moles of } (NH\u2084)\u2083PO\u2084 = \\frac{2267.96 \\, \\text{g}}{149.086 \\, \\text{g\/mol}} = 15.21 \\, \\text{mol}<\/p>\n\n\n\n

          (c) How many grams of ammonia and phosphoric acid are required to produce 1.00 kg of (NH\u2084)\u2083PO\u2084 if the reaction has 100% yield?<\/h3>\n\n\n\n

          From the balanced equation: 3\u2009mol NH\u2083\u2009:\u20091\u2009mol H\u2083PO\u2084\u2009:\u20091\u2009mol (NH\u2084)\u2083PO\u20843 \\, \\text{mol NH\u2083} \\, : \\, 1 \\, \\text{mol H\u2083PO\u2084} \\, : \\, 1 \\, \\text{mol (NH\u2084)\u2083PO\u2084}<\/p>\n\n\n\n

            \n
          1. Moles of (NH\u2084)\u2083PO\u2084:<\/strong> 1.00\u2009kg=1000\u2009g1.00 \\, \\text{kg} = 1000 \\, \\text{g} Moles\u00a0of\u00a0(NH\u2084)\u2083PO\u2084=1000\u2009g149.086\u2009g\/mol=6.71\u2009mol\\text{Moles of (NH\u2084)\u2083PO\u2084} = \\frac{1000 \\, \\text{g}}{149.086 \\, \\text{g\/mol}} = 6.71 \\, \\text{mol}<\/li>\n\n\n\n
          2. Moles of ammonia (NH\u2083):<\/strong> 3\u2009mol\u00a0NH\u2083\u2009:\u20091\u2009mol\u00a0(NH\u2084)\u2083PO\u20843 \\, \\text{mol NH\u2083} \\, : \\, 1 \\, \\text{mol (NH\u2084)\u2083PO\u2084} Moles\u00a0of\u00a0NH\u2083=3\u00d76.71=20.13\u2009mol\\text{Moles of NH\u2083} = 3 \\times 6.71 = 20.13 \\, \\text{mol} Grams of ammonia: 20.13\u2009mol\u00d717.034\u2009g\/mol=342.35\u2009g20.13 \\, \\text{mol} \\times 17.034 \\, \\text{g\/mol} = 342.35 \\, \\text{g}<\/li>\n\n\n\n
          3. Moles of phosphoric acid (H\u2083PO\u2084):<\/strong> Moles\u00a0of\u00a0H\u2083PO\u2084=6.71\u2009mol\\text{Moles of H\u2083PO\u2084} = 6.71 \\, \\text{mol} Grams of phosphoric acid: 6.71\u2009mol\u00d797.994\u2009g\/mol=657.39\u2009g6.71 \\, \\text{mol} \\times 97.994 \\, \\text{g\/mol} = 657.39 \\, \\text{g}<\/li>\n<\/ol>\n\n\n\n

            (d) How many kilograms (kg) of ammonia and phosphoric acid are required to produce 1.00 kg of (NH\u2084)\u2083PO\u2084 if the reaction has 85.0% yield?<\/h3>\n\n\n\n

            Theoretical amounts are calculated in part (c), but now we account for the 85% yield.<\/p>\n\n\n\n

              \n
            1. Ammonia required:<\/strong> 342.35\u2009g\u00f70.85=402.94\u2009g342.35 \\, \\text{g} \\div 0.85 = 402.94 \\, \\text{g} Kilograms\u00a0of\u00a0ammonia=402.94\u2009g\u00f71000=0.403\u2009kg\\text{Kilograms of ammonia} = 402.94 \\, \\text{g} \\div 1000 = 0.403 \\, \\text{kg}<\/li>\n\n\n\n
            2. Phosphoric acid required:<\/strong> 657.39\u2009g\u00f70.85=773.99\u2009g657.39 \\, \\text{g} \\div 0.85 = 773.99 \\, \\text{g} Kilograms\u00a0of\u00a0phosphoric\u00a0acid=773.99\u2009g\u00f71000=0.774\u2009kg\\text{Kilograms of phosphoric acid} = 773.99 \\, \\text{g} \\div 1000 = 0.774 \\, \\text{kg}<\/li>\n<\/ol>\n\n\n\n

              Summary:<\/h3>\n\n\n\n
                \n
              • (a) Molar masses: NH\u2083 = 17.034 g\/mol, H\u2083PO\u2084 = 97.994 g\/mol, (NH\u2084)\u2083PO\u2084 = 149.086 g\/mol.<\/li>\n\n\n\n
              • (b) Moles of ammonium phosphate in 5.00 lb: 15.21 mol.<\/li>\n\n\n\n
              • (c) Grams required for 1.00 kg of (NH\u2084)\u2083PO\u2084 (100% yield): 342.35 g of NH\u2083 and 657.39 g of H\u2083PO\u2084.<\/li>\n\n\n\n
              • (d) Grams required for 1.00 kg of (NH\u2084)\u2083PO\u2084 (85% yield): 0.403 kg of NH\u2083 and 0.774 kg of H\u2083PO\u2084.<\/li>\n<\/ul>\n\n\n\n

                I\u2019ll now generate the image of the chemical equation for you!<\/p>\n\n\n\n

                Here is the image of the balanced chemical equation for the preparation of ammonium phosphate. If you need further explanations or assistance, feel free to ask!<\/p>\n\n\n\n

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

                Ammonium phosphate fertilizer is prepared by the following reaction: 3NH3(g) + H3PO4(aq) \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae (NH4)3PO4(s) (a) \u00c3\u201a\u00c2 Calculate the molar mass of ammonia, phosphoric acid and ammonium phosphate. (b) \u00c3\u201a\u00c2 How many moles of ammonium phosphate are present in a 5.00-pounds bag of (NH4)3PO4?Ammonium phosphate fertilizer is prepared by the following reaction: 3NH3(g) + H3PO4(aq) \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae […]<\/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-192513","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192513","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=192513"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192513\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192513"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192513"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192513"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}