{"id":218779,"date":"2025-05-24T12:41:16","date_gmt":"2025-05-24T12:41:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218779"},"modified":"2025-05-24T12:41:18","modified_gmt":"2025-05-24T12:41:18","slug":"cisplatin-ptcl2nh32-is-a-drug-used-in-the-treatment-of-cancer","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/cisplatin-ptcl2nh32-is-a-drug-used-in-the-treatment-of-cancer\/","title":{"rendered":"Cisplatin, PtCl2(NH3)2, is a drug used in the treatment of cancer"},"content":{"rendered":"\n

Cisplatin, PtCl2(NH3)2, ?is a drug used in the treatment of cancer. It can be synthesized by the reaction of K2PtCl4 ?with ammonia. What mass of ammonia in grams is needed to produce 1.00kg ?of cisplatin?<\/p>\n\n\n\n

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

To determine the mass of ammonia (NH\u2083)<\/strong> needed to produce 1.00 kg (1000 g)<\/strong> of cisplatin (PtCl\u2082(NH\u2083)\u2082)<\/strong>, we must first use stoichiometry based on the balanced chemical reaction and molar masses.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Balanced Chemical Equation<\/strong><\/h3>\n\n\n\n

The synthesis of cisplatin from potassium tetrachloroplatinate(II) and ammonia is: K2PtCl4+2NH3\u2192PtCl2(NH3)2+2KCl\\text{K}_2\\text{PtCl}_4 + 2\\text{NH}_3 \\rightarrow \\text{PtCl}_2(\\text{NH}_3)_2 + 2\\text{KCl}<\/p>\n\n\n\n

From the equation, 2 moles of NH\u2083<\/strong> are required to form 1 mole of cisplatin<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Molar Masses<\/strong><\/h3>\n\n\n\n
    \n
  • PtCl\u2082(NH\u2083)\u2082 (cisplatin):<\/strong>\n
      \n
    • Pt = 195.08 g\/mol<\/li>\n\n\n\n
    • Cl\u2082 = 2 \u00d7 35.45 = 70.90 g\/mol<\/li>\n\n\n\n
    • NH\u2083 = 2 \u00d7 (14.01 + 3 \u00d7 1.008) = 2 \u00d7 17.034 = 34.068 g\/mol<\/li>\n\n\n\n
    • Total = 195.08 + 70.90 + 34.068 = 300.05 g\/mol<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • NH\u2083:<\/strong>\n
        \n
      • 14.01 + (3 \u00d7 1.008) = 17.034 g\/mol<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Step 3: Calculate Moles of Cisplatin<\/strong><\/h3>\n\n\n\n

        Moles of cisplatin=1000\u2009g300.05\u2009g\/mol\u22483.33\u2009mol\\text{Moles of cisplatin} = \\frac{1000\\, \\text{g}}{300.05\\, \\text{g\/mol}} \\approx 3.33\\, \\text{mol}<\/p>\n\n\n\n

        \n\n\n\n

        Step 4: Use Mole Ratio to Get Moles of NH\u2083<\/strong><\/h3>\n\n\n\n

        From the balanced equation: 1\u2009mol cisplatin\u21d22\u2009mol NH\u20831\\, \\text{mol cisplatin} \\Rightarrow 2\\, \\text{mol NH\u2083} 3.33\u2009mol cisplatin\u21d22\u00d73.33=6.66\u2009mol NH\u20833.33\\, \\text{mol cisplatin} \\Rightarrow 2 \\times 3.33 = 6.66\\, \\text{mol NH\u2083}<\/p>\n\n\n\n

        \n\n\n\n

        Step 5: Convert Moles of NH\u2083 to Grams<\/strong><\/h3>\n\n\n\n

        Mass NH\u2083=6.66\u2009mol\u00d717.034\u2009g\/mol\u2248113.5\u2009g\\text{Mass NH\u2083} = 6.66\\, \\text{mol} \\times 17.034\\, \\text{g\/mol} \\approx 113.5\\, \\text{g}<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Final Answer: 113.5 grams of ammonia are needed to produce 1.00 kg of cisplatin.<\/strong><\/h3>\n\n\n\n
        \n\n\n\n

        Explanation <\/strong><\/h3>\n\n\n\n

        Cisplatin, PtCl\u2082(NH\u2083)\u2082, is a platinum-based chemotherapy drug that works by binding to DNA in cancer cells, disrupting their function and leading to cell death. To synthesize cisplatin in the lab or industrially, potassium tetrachloroplatinate(II) (K\u2082PtCl\u2084) is reacted with ammonia (NH\u2083). The stoichiometry of the reaction reveals a 1:2 molar ratio between cisplatin and ammonia: one mole of cisplatin forms from two moles of ammonia.<\/p>\n\n\n\n

        To determine how much ammonia is required to produce 1.00 kg (1000 g) of cisplatin, we first calculate the number of moles of cisplatin needed. Using its molar mass (300.05 g\/mol), we find that 1000 g corresponds to approximately 3.33 moles of cisplatin. Given the 2:1 ratio, we need 6.66 moles of ammonia.<\/p>\n\n\n\n

        Next, we convert these moles into grams using the molar mass of ammonia, which is approximately 17.034 g\/mol. Multiplying 6.66 moles by this value yields roughly 113.5 grams of ammonia.<\/p>\n\n\n\n

        This calculation is essential in synthetic chemistry and pharmaceutical manufacturing to ensure efficient resource use and avoid excess reactants. The precise measurement of reactants also minimizes waste and cost while optimizing product yield and purity\u2014crucial factors in the production of high-quality drugs like cisplatin.<\/p>\n\n\n\n

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

        Cisplatin, PtCl2(NH3)2, ?is a drug used in the treatment of cancer. It can be synthesized by the reaction of K2PtCl4 ?with ammonia. What mass of ammonia in grams is needed to produce 1.00kg ?of cisplatin? The Correct Answer and Explanation is: To determine the mass of ammonia (NH\u2083) needed to produce 1.00 kg (1000 g) […]<\/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-218779","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218779","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=218779"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218779\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218779"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218779"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}