Answer to Question #115100 in Chemistry for Sara

Question

A sample (m = 0.6914 g) containing only MgSO4 and K2SO4 is treated with a barium nitrate-solution. Formulate the reaction equation(s) for this procedure. Calculate the percentage of each salt in the mixture, when 0.9797 g of dried precipitate were obtained?

Answer 1

MgSO₄ + Ba(NO₃)₂ = BaSO₄ + Mg(NO3)2.

K₂SO₄ + Ba(NO₃)₂ = BaSO₄ + 2KNO₃

0.9797/261=

120x + 174y = 0.6914.

x + y = 0,00375363985.

x=0,000708641963.

y= 0,00304499789.

120 x 0,000708641963 = 0,0850370356 g of MgSO_4.

174 x 0,00304499789 = 0,606362964 g of K₂SO₄

0,606362964 x 100 % / 0.6914 = 87,7 %. of K₂SO_4.

100 - 87.7 = 12.3 % of MgSO_4.