Question

A 355 mL sample of a HCl solution reacts with excess My to produce 4.20 L of H2 gas measured at 745 mmHg and 35 degrees Celsius. What is the molarity of the HCl solution?

Answer 1

Mg(s) + 2HCl(aq) = MgCl2(aq) + H2(g)

35 degrees Celsius=308.15

$R = \frac{PV}{nT}= \frac{760mmHg\times22400mL} {1mole\times 273K} = 62400mmHgmL/moleK$

$n = \frac{PV}{RT }=\frac{ 745mmHg\times355mL}{62400mmHgmL\times 308.15K} = 0.01376molesH2 =$

$0.01376molesH\times2=0.0275 moles H2$

$molarity=\frac{n}{V}= \frac{0.0275moles H}{0.355L} = 0.0775 moles /L$