A 0.160-g sample of unknown metal (X) reacts with hydrochloric acid to produce 66.5 mL of "wet" gas at 20 °C and 754 mm Hg. What is the unknown metal (X)? (The vapor pressure of water at 20 °C is 18 mm Hg. Help please?

Answer 1

I'm not sure this has a solution since there is no way to know the mole ratio between H2 produced and mols M consumed. I have assumed +2 metal and 1mol M to 1 mol H2 ratio to make MCl2 but it may not be that at all.
M + 2HCl ==> MCl2 + H2

PV = nRT
P = Ptotal - pH2O = 754-18 = 736
so P = 736/760 = about 0.968 atm.
V is 0.0665L
R you know
T = 20 + 273 = ?
Solve for n and I get approx 0.003.
Then 0.160/0.003 = approx ? You need to redo all of the math.