{"id":186835,"date":"2025-01-26T09:23:29","date_gmt":"2025-01-26T09:23:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186835"},"modified":"2025-01-26T09:23:31","modified_gmt":"2025-01-26T09:23:31","slug":"calculate-the-theoretical-yield-in-gram","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/26\/calculate-the-theoretical-yield-in-gram\/","title":{"rendered":"Calculate the theoretical yield in gram"},"content":{"rendered":"\n

Calculate the theoretical yield, in grams, of Eauropium Hydrogen phosphate (1 EuHPOsn formed rom 108.34-grams of Europium nitrate and 122.43-grams of Lithium Hydrogenphosphate. Eu(NO) + 3 Li2HPO4 I Eu2(HPO4)3 + 6 L\u0130NO3 13 pts)One of the compounds iridium and oxygen found to contain 88.90% iridium and 11.10 % What is the empirical formula of compound?<\/p>\n\n\n\n

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

Theoretical Yield of Europium Hydrogenphosphate (EuHPO\u2084)<\/strong><\/h3>\n\n\n\n

Let’s start by calculating the theoretical yield of Europium Hydrogenphosphate (EuHPO\u2084) from Europium Nitrate (Eu(NO\u2083)\u2083) and Lithium Hydrogenphosphate (Li\u2082HPO\u2084).<\/p>\n\n\n\n

Step 1: Write the balanced chemical equation<\/h4>\n\n\n\n

The balanced chemical equation is: Eu(NO\u2083)\u2083+3Li\u2082HPO\u2084\u2192Eu\u2082(HPO\u2084)\u2083+6LiNO\u2083\\text{Eu(NO\u2083)\u2083} + 3 \\text{Li\u2082HPO\u2084} \\rightarrow \\text{Eu\u2082(HPO\u2084)\u2083} + 6 \\text{LiNO\u2083}<\/p>\n\n\n\n

Step 2: Calculate the molar masses of reactants<\/h4>\n\n\n\n
    \n
  • Eu(N\u2083)\u2083<\/strong>:\n
      \n
    • Eu: 151.98 g\/mol<\/li>\n\n\n\n
    • N: 14.01 g\/mol (\u00d73)<\/li>\n\n\n\n
    • O: 16.00 g\/mol (\u00d79)<\/li>\n\n\n\n
    • Molar mass of Eu(NO\u2083)\u2083 = 151.98 + (3 \u00d7 14.01) + (9 \u00d7 16.00) = 151.98 + 42.03 + 144.00 = 338.01 g\/mol<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Li\u2082HPO\u2084<\/strong>:\n
        \n
      • Li: 6.94 g\/mol (\u00d72)<\/li>\n\n\n\n
      • H: 1.008 g\/mol (\u00d71)<\/li>\n\n\n\n
      • P: 30.97 g\/mol (\u00d71)<\/li>\n\n\n\n
      • O: 16.00 g\/mol (\u00d74)<\/li>\n\n\n\n
      • Molar mass of Li\u2082HPO\u2084 = (2 \u00d7 6.94) + 1.008 + 30.97 + (4 \u00d7 16.00) = 13.88 + 1.008 + 30.97 + 64.00 = 109.86 g\/mol<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

        Step 3: Convert the mass of the reactants to moles<\/h4>\n\n\n\n
          \n
        • Moles of Eu(NO\u2083)\u2083: 108.34\u2009g338.01\u2009g\/mol=0.3204\u2009mol\\frac{108.34 \\, \\text{g}}{338.01 \\, \\text{g\/mol}} = 0.3204 \\, \\text{mol}<\/li>\n\n\n\n
        • Moles of Li\u2082HPO\u2084: 122.43\u2009g109.86\u2009g\/mol=1.113\u2009mol\\frac{122.43 \\, \\text{g}}{109.86 \\, \\text{g\/mol}} = 1.113 \\, \\text{mol}<\/li>\n<\/ul>\n\n\n\n

          Step 4: Determine the limiting reagent<\/h4>\n\n\n\n

          From the balanced equation, 1 mole of Eu(NO\u2083)\u2083 reacts with 3 moles of Li\u2082HPO\u2084. Therefore:<\/p>\n\n\n\n

            \n
          • 0.3204 moles of Eu(NO\u2083)\u2083 would require: 0.3204\u2009mol\u00d73=0.9612\u2009mol\u00a0of\u00a0Li\u2082HPO\u20840.3204 \\, \\text{mol} \\times 3 = 0.9612 \\, \\text{mol of Li\u2082HPO\u2084}<\/li>\n<\/ul>\n\n\n\n

            Since we have 1.113 moles of Li\u2082HPO\u2084 (which is more than required), Eu(NO\u2083)\u2083<\/strong> is the limiting reagent.<\/p>\n\n\n\n

            Step 5: Calculate the moles of Eu\u2082(HPO\u2084)\u2083 formed<\/h4>\n\n\n\n

            From the balanced equation, 1 mole of Eu(NO\u2083)\u2083 produces 1 mole of Eu\u2082(HPO\u2084)\u2083. Therefore, the moles of Eu\u2082(HPO\u2084)\u2083 formed are 0.3204 mol.<\/p>\n\n\n\n

            Step 6: Calculate the theoretical yield of Eu\u2082(HPO\u2084)\u2083<\/h4>\n\n\n\n
              \n
            • Molar mass of Eu\u2082(HPO\u2084)\u2083:\n
                \n
              • Eu: 151.98 g\/mol (\u00d72)<\/li>\n\n\n\n
              • H: 1.008 g\/mol (\u00d73)<\/li>\n\n\n\n
              • P: 30.97 g\/mol (\u00d73)<\/li>\n\n\n\n
              • O: 16.00 g\/mol (\u00d712)<\/li>\n\n\n\n
              • Molar mass of Eu\u2082(HPO\u2084)\u2083 = (2 \u00d7 151.98) + (3 \u00d7 1.008) + (3 \u00d7 30.97) + (12 \u00d7 16.00) = 303.96 + 3.024 + 92.91 + 192.00 = 591.89 g\/mol<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

                The theoretical yield of Eu\u2082(HPO\u2084)\u2083 is: 0.3204\u2009mol\u00d7591.89\u2009g\/mol=189.78\u2009g0.3204 \\, \\text{mol} \\times 591.89 \\, \\text{g\/mol} = 189.78 \\, \\text{g}<\/p>\n\n\n\n

                So, the theoretical yield of Eu\u2082(HPO\u2084)\u2083 is 189.78 g<\/strong>.<\/p>\n\n\n\n

                \n\n\n\n

                Empirical Formula of the Compound Containing Iridium and Oxygen<\/strong><\/h3>\n\n\n\n

                Given that the compound contains 88.90% iridium (Ir)<\/strong> and 11.10% oxygen (O)<\/strong>, let’s calculate the empirical formula.<\/p>\n\n\n\n

                Step 1: Assume 100 grams of the compound<\/h4>\n\n\n\n
                  \n
                • Mass of Iridium = 88.90 g<\/li>\n\n\n\n
                • Mass of Oxygen = 11.10 g<\/li>\n<\/ul>\n\n\n\n

                  Step 2: Convert masses to moles<\/h4>\n\n\n\n
                    \n
                  • Moles of Iridium: 88.90\u2009g192.22\u2009g\/mol=0.463\u2009mol\u00a0of\u00a0Ir\\frac{88.90 \\, \\text{g}}{192.22 \\, \\text{g\/mol}} = 0.463 \\, \\text{mol of Ir}<\/li>\n\n\n\n
                  • Moles of Oxygen: 11.10\u2009g16.00\u2009g\/mol=0.694\u2009mol\u00a0of\u00a0O\\frac{11.10 \\, \\text{g}}{16.00 \\, \\text{g\/mol}} = 0.694 \\, \\text{mol of O}<\/li>\n<\/ul>\n\n\n\n

                    Step 3: Determine the simplest ratio<\/h4>\n\n\n\n

                    To find the ratio of Ir to O, divide each by the smaller number of moles (0.463):<\/p>\n\n\n\n

                      \n
                    • Ratio of Iridium: 0.4630.463=1\\frac{0.463}{0.463} = 1<\/li>\n\n\n\n
                    • Ratio of Oxygen: 0.6940.463\u22481.5\\frac{0.694}{0.463} \\approx 1.5<\/li>\n<\/ul>\n\n\n\n

                      Since the ratio of O is approximately 1.5, we multiply both by 2 to get whole numbers:<\/p>\n\n\n\n

                        \n
                      • Iridium: 1\u00d72=21 \\times 2 = 2<\/li>\n\n\n\n
                      • Oxygen: 1.5\u00d72=31.5 \\times 2 = 3<\/li>\n<\/ul>\n\n\n\n

                        Thus, the empirical formula of the compound is Ir\u2082O\u2083<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

                        Calculate the theoretical yield, in grams, of Eauropium Hydrogen phosphate (1 EuHPOsn formed rom 108.34-grams of Europium nitrate and 122.43-grams of Lithium Hydrogenphosphate. Eu(NO) + 3 Li2HPO4 I Eu2(HPO4)3 + 6 L\u0130NO3 13 pts)One of the compounds iridium and oxygen found to contain 88.90% iridium and 11.10 % What is the empirical formula of compound? […]<\/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-186835","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186835","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=186835"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186835\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186835"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186835"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186835"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}