Methanol is manufactured by the reaction of carbon monoxide gas with hydrogen gas. If a $95~\%$ yield is usually obtained, approximately how many liters of hydrogen gas at $\pu{350 °C}$ and $\pu{300 atm}$ are usually required to make $1.0$ liter of methanol measured at $\pu{20.0 °C}?$ The density of methanol at $\pu{20.0 °C}$ is $\pu{0.8 g mL^-1}.$
(a) $\pu{0.89 L}$
(b) $\pu{2.5 L}$
(c) $\pu{4.8 L}$
(d) $\pu{5.0 L}$
(e) $\pu{8.1 L}$
The correct answer is (d) $\pu{5.0 L}.$
I got two different answers, both wrong. I have detailed them below.
Attempt 1
I found the mass of methanol:
$$m(\ce{MeOH}) = \rho V = (\pu{0.8 g mL^-1})(\pu{1000 mL}) = \pu{800 g},\tag{1.1}$$
and calculated its theoretical yield:
$$m_\mathrm{theor}(\ce{MeOH}) = 0.95\times\pu{800 g} = \pu{842 g}.\tag{1.2}$$
The amount of methanol is
$$n(\ce{MeOH}) = \frac{\pu{842 g}}{\pu{32 g mol^-1}} = \pu{26.3 mol}.\tag{1.3}$$
The amount of hydrogen from balanced equation is $n(\ce{H2}) = \pu{52.6 mol}.$ From here
$$ \begin{align} V(\ce{H2}) &= \frac{nRT}{p} \\ &= \frac{(\pu{52.6 mol})(\pu{8.31 J mol^-1 K^-1})(\pu{(350 + 273) K})}{\pu{3E7 Pa}} \\ &= \pu{9.08E-3 m^3} \\ &\approx \pu{9.1 L}. \tag{1.4} \end{align} $$
Attempt 2
Assuming the pressure of methanol is $\pu{1 atm} = \pu{1E5 Pa},$
$$ \begin{align} n &= \frac{pV}{RT} \\ &= \frac{(\pu{1E5 Pa})(\pu{1000 mL})}{(\pu{8.31 J mol^-1 K^-1})(\pu{293 K})} \\ &= \pu{41071 mol}. \tag{2.1} \end{align} $$
Theoretical amount of methanol is
$$n_\mathrm{theor}(\ce{MeOH}) = \pu{41071 mol}/0.95 = \pu{43232 mol}.\tag{2.2}$$
The amount of hydrogen from the balanced equation is $n(\ce{H2}) = \pu{86464 mol}.$ From here
$$ \begin{align} V(\ce{H2}) &= \frac{nRT}{p} \\ &= \frac{(\pu{86464 mol})(\pu{8.31 J mol^-1 K^-1})(\pu{(350 + 273) K})}{\pu{3E7 Pa}} \\ &= \pu{14.92 m^3}. \tag{2.3} \end{align} $$
Attempt 2 is definitely wrong. I don't get what I did wrong in my attempt 1. What am I missing?
