Chapter 8 - Properties of Gases - Questions for Review and Thought - Topical Questions - Page 370b: 50

4.51 atm

$n(H_{2})=\frac{1.50\,g}{2.016\,g/mol}=0.744\,mol$ $n(N_{2})= \frac{5.00\,g}{28.0134\,g/mol}=0.178\,mol$ $n_{total}=n(H_{2})+n(N_{2})=(0.744+0.178)mol=0.922\,mol$ $P_{total}=\frac{n_{total}RT}{V}$ $=\frac{(0.922\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(298\,K)}{5.0\,L}$ $=4.51\,atm$