{"id":218776,"date":"2025-05-24T12:36:03","date_gmt":"2025-05-24T12:36:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218776"},"modified":"2025-05-24T12:36:05","modified_gmt":"2025-05-24T12:36:05","slug":"cisplatin-ptnh32cl2-an-anticancer-agent-used-for-the-treatment-of-solid-tumors","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/cisplatin-ptnh32cl2-an-anticancer-agent-used-for-the-treatment-of-solid-tumors\/","title":{"rendered":"Cisplatin [Pt(NH3)2Cl2], an anticancer agent used for the treatment of solid tumors"},"content":{"rendered":"\n

Cisplatin [Pt(NH3)2Cl2], an anticancer agent used for the treatment of solid tumors, is prepared by the reaction of ammonia (NH3) with potassium tetrachloroplatinate (K2PtCl4). Assume that 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react. K2PtCl4 + 2 NH3 \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae Pt(NH3)2Cl2 + 2 KCl a) Which reactant is limiting, and which is in excess? b) How many grams of the excess reactant are consumed, and how many grams remain? c) How many grams of cisplatin are formed?
Cisplatin [Pt(NH3)2Cl2], an anticancer agent used for the treatment of
solid tumors, is prepared by the reaction of ammonia (NH3) with potassium
tetrachloroplatinate (K2PtCl4). Assume that 10.0 g of K2PtCl4 and 10.0 g of NH3 are
allowed to react.
K2PtCl4 + 2 NH3 \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae Pt(NH3)2Cl2 + 2 KCl
a) Which reactant is limiting, and which is in excess?
b) How many grams of the excess reactant are consumed, and how many grams remain?
c) How many grams of cisplatin are formed?<\/p>\n\n\n\n

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

Balanced Chemical Equation:<\/strong><\/h3>\n\n\n\n

K2PtCl4+2NH3\u2192Pt(NH3)2Cl2+2KCl\\text{K}_2\\text{PtCl}_4 + 2 \\text{NH}_3 \\rightarrow \\text{Pt(NH}_3)_2\\text{Cl}_2 + 2 \\text{KCl}<\/p>\n\n\n\n

\n\n\n\n

Molar Masses:<\/strong><\/h3>\n\n\n\n
    \n
  • K\u2082PtCl\u2084 = 2(39.1) + 195.1 + 4(35.5) = 415.1 g\/mol<\/strong><\/li>\n\n\n\n
  • NH\u2083 = 14.0 + 3(1.0) = 17.0 g\/mol<\/strong><\/li>\n\n\n\n
  • Cisplatin [Pt(NH\u2083)\u2082Cl\u2082] = 195.1 + 2(17.0) + 2(35.5) = 300.1 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Moles of Each Reactant:<\/strong><\/h3>\n\n\n\n
      \n
    • K\u2082PtCl\u2084:
      10.0\u00a0g415.1\u00a0g\/mol\u22480.0241\u00a0mol\\frac{10.0 \\text{ g}}{415.1 \\text{ g\/mol}} \\approx 0.0241 \\text{ mol}<\/li>\n\n\n\n
    • NH\u2083:
      10.0\u00a0g17.0\u00a0g\/mol\u22480.588\u00a0mol\\frac{10.0 \\text{ g}}{17.0 \\text{ g\/mol}} \\approx 0.588 \\text{ mol}<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Limiting Reactant:<\/strong><\/h3>\n\n\n\n

      From the balanced equation:<\/p>\n\n\n\n

        \n
      • 1 mol of K\u2082PtCl\u2084 reacts with 2 mol of NH\u2083.<\/li>\n\n\n\n
      • For 0.0241 mol of K\u2082PtCl\u2084, we need:
        0.0241\u00d72=0.0482\u00a0mol\u00a0NH30.0241 \\times 2 = 0.0482 \\text{ mol NH}_3<\/li>\n<\/ul>\n\n\n\n

        We have 0.588 mol NH\u2083, which is much more<\/strong> than needed.<\/p>\n\n\n\n

        \u2705 K\u2082PtCl\u2084 is the limiting reactant<\/strong>
        \u2705 NH\u2083 is in excess<\/strong><\/p>\n\n\n\n

        \n\n\n\n

        (a) Answer:<\/strong><\/h3>\n\n\n\n
          \n
        • Limiting reactant<\/strong>: K\u2082PtCl\u2084<\/li>\n\n\n\n
        • Excess reactant<\/strong>: NH\u2083<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          (b) Grams of NH\u2083 Consumed and Remaining:<\/strong><\/h3>\n\n\n\n
            \n
          • NH\u2083 consumed = 2 \u00d7 0.0241 mol = 0.0482 mol<\/li>\n\n\n\n
          • Grams consumed = 0.0482 mol \u00d7 17.0 g\/mol = 0.82 g<\/strong><\/li>\n\n\n\n
          • Grams remaining = 10.0 g \u2013 0.82 g = 9.18 g<\/strong><\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            (c) Grams of Cisplatin Formed:<\/strong><\/h3>\n\n\n\n
              \n
            • From 1 mol K\u2082PtCl\u2084 \u2192 1 mol cisplatin<\/li>\n\n\n\n
            • So, 0.0241 mol K\u2082PtCl\u2084 \u2192 0.0241 mol cisplatin<\/li>\n\n\n\n
            • Mass = 0.0241 mol \u00d7 300.1 g\/mol = 7.23 g<\/strong><\/li>\n<\/ul>\n\n\n\n
              \n\n\n\n

              Final Answers:<\/strong><\/h3>\n\n\n\n

              a)<\/strong> Limiting reactant: K\u2082PtCl\u2084<\/strong>, Excess: NH\u2083<\/strong>
              b)<\/strong> NH\u2083 consumed: 0.82 g<\/strong>, NH\u2083 remaining: 9.18 g<\/strong>
              c)<\/strong> Mass of cisplatin formed: 7.<\/strong><\/p>\n\n\n\n

              <\/h3>\n\n\n\n

              The preparation of cisplatin involves a simple stoichiometric reaction between potassium tetrachloroplatinate (K\u2082PtCl\u2084) and ammonia (NH\u2083). To determine the limiting reagent and calculate the product mass, we start by analyzing the chemical reaction: K2PtCl4+2NH3\u2192Pt(NH3)2Cl2+2KCl\\text{K}_2\\text{PtCl}_4 + 2 \\text{NH}_3 \\rightarrow \\text{Pt(NH}_3)_2\\text{Cl}_2 + 2 \\text{KCl}<\/p>\n\n\n\n

              This equation shows that 1 mole of K\u2082PtCl\u2084 reacts with 2 moles of NH\u2083 to produce 1 mole of cisplatin. The first step is to convert the given masses into moles using their molar masses. For K\u2082PtCl\u2084, we calculate about 0.0241 mol, and for NH\u2083, we have about 0.588 mol.<\/p>\n\n\n\n

              Using stoichiometry, we find that the amount of NH\u2083 needed for the reaction is 0.0482 mol (2 \u00d7 0.0241 mol). Since 0.588 mol is available, which is significantly more than needed, NH\u2083 is the excess reactant. This makes K\u2082PtCl\u2084 the limiting reactant, as it restricts the amount of product that can be formed.<\/p>\n\n\n\n

              Next, we determine the mass of NH\u2083 consumed: 0.0482 mol \u00d7 17.0 g\/mol = 0.82 g. Therefore, 10.0 g \u2013 0.82 g = 9.18 g NH\u2083 remains unreacted.<\/p>\n\n\n\n

              Finally, using the 1:1 molar ratio between K\u2082PtCl\u2084 and cisplatin, we find that 0.0241 mol of K\u2082PtCl\u2084 will produce 0.0241 mol of cisplatin. This equates to a mass of 7.23 g.<\/p>\n\n\n\n

              This stoichiometric analysis is crucial in synthetic chemistry and pharmaceutical manufacturing, ensuring efficient resource use and accurate prediction of product yield.<\/p>\n\n\n\n

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

              Cisplatin [Pt(NH3)2Cl2], an anticancer agent used for the treatment of solid tumors, is prepared by the reaction of ammonia (NH3) with potassium tetrachloroplatinate (K2PtCl4). Assume that 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react. K2PtCl4 + 2 NH3 \u00c3\u00af\u00e2\u20ac\u0161\u00c2\u00ae Pt(NH3)2Cl2 + 2 KCl a) Which reactant is limiting, and which […]<\/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-218776","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218776","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=218776"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218776\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218776"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218776"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218776"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}