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?

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?

I was given this question.

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 350 °C and 300 atm are usually required to make 1.0 liter of methanol measured at 20.0 °C. The density of methanol at 20.0 °C is 0.8 g/mL. (a) 0.89 L (b) 2.5 L (c) 4.8 L (d) 5.0 L (e) 8.1 L

The correct answer is D) 5.0 L according to the mark scheme. How can I get this answer?

I got two different answers, both wrong. I have detailed them below.

Attempt 1: mass of methanol = D x v = 0.8 x 1000 = 800 g % yield = (800/T.Y) x 100 = 95 Theoretical Yield (T.Y) of methanol = 842.1 Moles of methanol = 842.1/32 = 26.3 mol Moles of hydrogen from balanced equation = 52.63 mol Using V=nRT/P Volume of hydrogen = (52.63x8.31x(350+273))/(300x10^5) = 9.08x10^-3 m^3 = 9.08 dm^3

Attempt 2: Assuming the pressure of methanol is 1 atm = 1x10^5 Pa Use n=PV/RT=(1x10^5x1000)/(8.31x293)= 41071 mol % yield = (41071/T.Y) x 100 = 95 Theoretical no of moles (T.Y) of methanol = 43232 mol Moles of hydrogen from balanced equation = 86464.4 mol Using V=nRT/P Volume of hydrogen = (86464.4x8.31x(350+273))/(300x10^5) = 14.92 m^3

Attempt 2 is definitely wrong. I don't get what I did wrong in my attempt 1. What am I missing? What is the calculation to get this answer?

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?