## Chemistry: The Molecular Science (5th Edition)

Moles of $B_{4}H_{10}$= $\frac{mass\,of\,B_{4}H_{10}}{molar\,mass\,of\,B_{4}H_{10}}=\frac{0.05\,g}{53.32\,g/mol}$ $=0.0009377 mol$ When 2 moles of $B_{4}H_{10}$ are reacted, according to the equation, 10 moles of $H_{2}O$ are produced. 1 mol of $B_{4}H_{10}$ produces $\frac{10}{2}=5$ moles of $H_{2}O$. $\implies$ moles of $H_{2}O$ produced (n)$=5\times0.0009377\,mol=0.0046885\,mol$ $V=4.25\,L$ $T=(30+273)K= 303\,K$ $PV=nRT$ (ideal gas law) $\implies P= \frac{nRT}{V}=\frac{(0.0046885\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(303\,K)}{4.25\,L}$ $=0.03\,atm$