How many grams of Ca was contained in a mixture of calcium and magnesium if, upon interaction of 16 grams

How many grams of Ca was contained in a mixture of calcium and magnesium if, upon interaction of 16 grams of this mixture with an excess of hydrochloric acid, 0.6 mol of H2 was released. Ar (Ca) = 40: Ar (Mg) = 24

Given:
m mixture (Ca, Mg) = 16 g
n (H2) = 0.6 mol

To find:
m (Ca) -?

1) Ca + 2HCl => CaCl2 + H2 ↑;
Mg + 2HCl => MgCl2 + H2 ↑;
2) Let n (Ca) = (x) mol, and n (Mg) = (y) mol;
3) n1 (H2) = n (Ca) = (x) mol;
4) n2 (H2) = n (Mg) = (y) mol;
5) n total (H2) = n1 (H2) + n2 (H2);
6) m (Ca) = n (Ca) * M (Ca) = (40x) g;
7) m (Mg) = n (Mg) * M (Mg) = (24y) g;
8) m mixture (Ca, Mg) = m (Ca) + m (Mg);
40x + 24y = 16;
x + y = 0.6;
y = 0.6 – x;
40x + 24 * (0.6 – x) = 16;
x = 0.1;
9) m (Ca) = 40x = 40 * 0.1 = 4 g.

Answer: Ca mass is 4 g.