Answer
$P_{H_{2}}=3.7\,atm$
$P_{Cl_{2}}=4.9\,atm$
Work Step by Step
$P_{total}V=n_{total}RT$
$\implies P_{total}=\frac{n_{total}RT}{V}$
$=\frac{[n(H_{2})+n(Cl_{2})]RT}{V}$
$=\frac{[\frac{3.0\,g}{2.016\,g/mol}+\frac{140.\,g}{70.9\,g/mol}](0.0821\,L\,atm\,mol^{-1}K^{-1})(301\,K)}{10.\,L}$
$=8.557076\,atm$
Partial pressures of $H_{2}$, $P_{H_{2}}=X_{H_{2}}P_{total}$:
$=\frac{\frac{3.0\,g}{2.016\,g/mol}}{\frac{3.0\,g}{2.016\,g/mol}+\frac{140.\,g}{70.9\,g/mol}}\times 8.557076\,atm$
$=3.7\,atm$
$P_{Cl_{2}}=X_{Cl_{2}}P_{total}$$=\frac{\frac{140.\,g}{70.9\,g/mol}}{\frac{3.0\,g}{2.016\,g/mol}+\frac{140.\,g}{70.9\,g/mol}}\times8.557076\,atm$
$=4.9\,atm$
