So if we use the equation:
[tex] CO+2H_{2} [/tex] → [tex] CH_{3}OH [/tex]
We can then determine the amount of [tex] H_{2} [/tex] needed to produce 208 kg of methanol.
So let's find out how many moles of methanol 208 kg is:
Methanol molar weight = 32.041g/mol
So then we can solve for moles of methanol:
[tex] 208kg*\frac{1,000g}{1kg} *\frac{1mol}{32.041g} =6,491.68mol [/tex]
So now that we have the amount of moles produced, we can use the molar ratio (from the balanced equation) of hydrogen and methanol. This ratio is 2:1 hydrogen:methanol.
Therefore, we can set up a proportion to solve for the moles of hydrogen needed:
[tex] \frac{2}{1} =\frac{x}{6,491.68} [/tex]
[tex] x=12,983.36mol [/tex]
So now that we have the number of moles of [tex] H_{2} [/tex] that are produced, we can then use the molar weight of hydrogen to solve for the mass that is needed:
[tex] 12,983.36mol*\frac{2.016g}{1mol} =26,174.45g_H_{2} [/tex]
Therefore, the amount of diatomic hydrogen ([tex] H_{2} [/tex]) that is needed to produce 208kg of methanol is [tex] 2.62x10^{4} [/tex]g.
