The volume of HCl gas required to react with excess magnesium metal to produce 6.82 L of hydrogen gas at 2.19 atm and 35.0 °C is ________ L.
A)6.82
B)2.19
C)13.6
D)4.38
E)3.41

Question

Quizplus · Accepted Answer

The balanced chemical equation is:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

From the ideal gas law, we have:

n(H2) = PV/(RT) = (2.19 atm)(6.82 L)/(0.0821 L·atm/mol·K)(308 K) = 0.634 mol H2

According to the balanced equation, 1 mole of magnesium reacts with 2 moles of HCl to produce 1 mole of hydrogen gas. Therefore, the number of moles of HCl required is:

n(HCl) = 0.634 mol H2 × (1 mol HCl)/(0.5 mol H2) = 1.268 mol HCl

From the ideal gas law again, we have:

PV = nRT

The conditions for the HCl gas are not given, but assuming they are the same as for hydrogen gas (which is typical), we can use the same values of P and T. Rearranging the equation to solve for V, we get:

V(HCl) = n(HCl)RT/P = (1.268 mol)(0.0821 L·atm/mol·K)(308 K)/(2.19 atm) = 13.6 L

Therefore, the volume of HCl gas required is 13.6 L. The best choice for this problem is (C).

The Volume of HCl Gas Required to React with Excess