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Answer in Organic Chemistry for M.saeed #122216

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April 21st, 2023

n(K2PtCl4) = 20 / 415 = 0.0482 mol

n(Pt(NH3)2Cl2) = 12.36 / 300 = 0.0412 mol

“alpha” = 0.0412 / 0.0482 × 100% = 85.5%

m2(NH3) = 0.0482 × 17 = 0.82 g

m2(Pt(NH3)2Cl2) = 0.0482 × 300 = 14.46 g

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