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A mixture of KBr and K2S of mass 15.7 g contains 8.36 g of potassium. What...

A mixture of KBr and K2S of mass 15.7 g contains 8.36 g of potassium. What is the percent by mass of KBr in the mixture?

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Answer #1

Let the mass of KBr = x gms

Mass of K2S = (15.7 - x )gms

Moles of KBr = mass /molar mass = x / 119

1 mole of KBr contains 1 mole of K

So

Moles of K = x / 119

Mass of K = moles*Molar mass = (x /119)*39 = 0.3277 x

Moles of K2S = (15.7 -x ) / 110.262

1 mole of K2S contains 2 moles of K

Moles of K = 2*(15.7 -x ) / 110.262 = 31.4 - 2x / 110.262

Mass of K =  [31.4 - 2x] * 39 / 110.262 = 1224.6 - 78x / 110.262 = 11.106 -0.7074x

Total mass of K = 8.36 gms

11.106 -0.7074x+ 0.3277 x = 8.36

x = 7.23

% mass of KBr = 7.23*100 / 15.7 = 46.063 %

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