A 300 ML sample of an HCL solution reacts with excess Mg to produce 4.10 Liter of H2 gas measured at 746 mmHg and 40 degrees Celsius.
Mg(S)+2HCL(aq)=MgCl2(aq)+H2(g).
What is the molarity of the HCL solution?

Question

A 300 ML sample of an HCL solution reacts with excess Mg to produce 4.10 Liter of H2 gas measured at 746 mmHg and 40 degrees Celsius. 
Mg(S)+2HCL(aq)=MgCl2(aq)+H2(g). 
What is the molarity of the HCL solution?

anonymous · Answer

We are given enough information to find the mols of H2 using the ideal gas law equation.
$$PV=nRT$$
$$\frac{ 746torr }{ 760torr }(4.10L)=n(0.0821\frac{ L*atm }{ mol*K })(313K)$$
$$n=0.15561 mols$$
n is our mols of H2 gas. Mg is excess, so HCL is our limiting agent. We can use stoichiometry to find the mols of HCL.
$$0.15661 mols H_2\left( \frac{ 2 mol HCL }{ 1 mol H_2 } \right)=0.31322 mols HCL$$
We have 300mL of HLC, and we know the mols of HCL, so we find the molarity.
$$Molarity=\frac{ mols }{ liters } \rightarrow Molarity=\frac{ 0.31322 mols HCL }{ 0.300 L HCL }=1.04M HCL$$