Answer
0.03 atm
Work Step by Step
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)
$\implies P= \frac{nRT}{V}=\frac{(0.0046885\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(303\,K)}{4.25\,L}$
$=0.03\,atm$
