Help - Chem?

Question

A 4.182 g sample of a mixture Zn and Al is treated with excess HCl. If 0.1686 moles of H2 are obtained, then what os the % by mass of Zn in the original mixture?

Answer 1

Use ALGEBRA!
Assume there is X grams of Zn in the original mixture.
X grams of Zn <==> X/65.41 mol of Zn
X grams of Zn <==> (4.182 - X)g Al <==> (4.182 - X)/26.98 mol of Al
0.1686 moles of H2 = 0.3372 mol H atoms ==>
0.3372 = 2*X/65.41 + 3*(4.182 - X)/26.98
Solve for X and calculate the requested value (X/4.182)*100%.
You do the math.