# Answer to Question #170652 in General Chemistry for uhhhh duhh

Question #170652

1. Given the following equation:

_____ K2PtCl4 + _____ NH3 g _____ Pt(NH3)2Cl2 + _____ KCl

a) Balance the equation.

b) Determine the theoretical yield of KCl if you start with 34.5 grams of NH3.

c) Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield?

1
2021-03-11T08:46:26-0500

a) K2PtCl4 + 2NH3 → Pt(NH3)2Cl2 + 2KCl

b) M(NH3) = 17.03 g/mol

n(NH3) "= \\frac{m}{M} = \\frac{34.5}{17.03} = 2.025 \\;mol"

According to the reaction equation:

n(KCl) = n(NH3) = 2.025 mol

M(KCl) = 74.55 g/mol

m(KCl) "= n \\times M = 2.025 \\times 74.55 = 150.96 \\;g"

The theoretical yield of KCl is 150.96 g.

c) n(NH3) "= \\frac{m}{M} = \\frac{34.5}{17.03} = 2.025 \\;mol"

According to the reaction equation:

n(Pt(NH3)2Cl2) = "\\frac{1}{2}" n(NH3) = 1.012 mol

M(Pt(NH3)2Cl2) = 301.1 g/mol

m(Pt(NH3)2Cl2) "= n \\times M = 1.012 \\times 301.1 = 304.71 \\;g"

Proportion:

304.71 g – 100 %

76.4 g – x

"x = \\frac{76.4 \\times 100}{304.71} = 25 \\%"

The percent yield is 25 %.

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