Antitumor agent, cisplatin, Pt(NH3)2Cl2 is prepared from the following reaction:

Question

Antitumor agent, cisplatin, Pt(NH3)2Cl2 is prepared from the following reaction: K2PtCl4 + 2 NH3 ? Pt(NH3)2Cl2 + 2 KCl Assume that 10.0 g of K2PtCl4 reacts with 10.0 g of NH3. (a) Which reactant is limiting and which is in excess? (b) How many grams of the excess reactant are consumed? (c) How many grams of cisplatin are formed?

Explanation / Answer

K2PtCl4 + 2 NH3 ---------> Pt(NH3)2Cl2 + 2 KCl 415.3 17 300.1 74.6 Mol. Wt a) # mol K2PtCl4 = 10.0g / 415.3 g.mol = 0.0241 mol-1 # mol NH3 = 10.0g / 17.0 g.mol = 0.588 mol / 2 = 0.294 mol -1 b) #mol = 0.0241 mol LR ( 2 NH3/1 K2PtCl4) = 0.0482 mol NH3 #g = 0.0482 mol * 17.0 g.mol-1 = 0.819 g USED c) # moles of Pt(NH3)2Cl2 = 0.0241 mol K2PtCl4 (1:1 ratio) = 0.0241 mol # g = 0.0241 mol * 300.1 g.mol-1 = 7.22 g of cisplatin d) %Yield = 6.53 g / 7.23 g = 0.9031 90.3% yield

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