# 11.6 g of a mixture of Na and K with H2SO4 formed 28.05 g of a mixture of Na and K sulphates.

11.6 g of a mixture of Na and K with H2SO4 formed 28.05 g of a mixture of Na and K sulphates. Determine the percentage of salts.

Given:
m (Na, K) = 11.6 g
m (Na2SO4, K2SO4) = 28.05 g

To find:
ω (Na2SO4) -?
ω (K2SO4) -?

1) 2Na + H2SO4 => Na2SO4 + H2;
2K + H2SO4 => K2SO4 + H2;
2) Let m (Na) = (x) r, then m (K) = (11.6 – x) r;
3) n (Na) = m / M = (x / 23) mol;
4) n (Na2SO4) = n (Na) = (x / 23) / 2 = (x / 46) mol;
5) m (Na2SO4) = n * M = (x / 46) * 142 = (142x / 46) g;
6) n (K) = m / M = ((11.6 – x) / 39) mol;
7) n (K2SO4) = n (K) = ((11.6 – x) / 39) / 2 = ((11.6 – x) / 78) mol;
8) m (K2SO4) = n * M = ((11.6 – x) / 78) * 174 = (174 * (11.6 – x) / 78) g;
9) (142x / 46) + (174 * (11.6 – x) / 78) = 28.05;
x = 2.538
10) m (Na2SO4) = 142x / 46 = 142 * 2.538 / 46 = 7.835 g;
11) ω (Na2SO4) = m * 100% / m mixture = 7.835 * 100% / 28.05 = 27.93%;
12) ω (K2SO4) = 100% – ω (Na2SO4) = 100% – 27.93% = 72.07%.

Answer: The mass fraction of Na2SO4 is 27.93%; K2SO4 – 72.07%.

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