Question

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?

Answer 1

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 %.