Question

Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) + 2H2(g) -> C H3OH(l)

Answer 1

Answer:

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.

Explanation:

The balanced reaction:

CO (g) + 2 H₂ (g) -> CH₃OH (l)

By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:

• CO: 1 mole
• H₂: 2 moles
• CH₃OH: 1 mole

Being the molar mass of each compound:

• CO: 28 g/mole
• H₂: 1 g/mole
• CH₃OH: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• CO: 1 mole* 28 g/mole= 28 grams
• H₂: 2 moles* 1 g/mole= 2 grams
• CH₃OH: 1 mole* 32 g/mole= 32 grams

Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:

• If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?

$mass of carbon monoxide=\frac{6000 grams of methanol*28 grams of carbon monoxide}{32 grams of methanol}$

mass of carbon monoxide= 5250 grams= 5.25 kg

If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?

$mass of hydrogen=\frac{6000 grams of methanol*2 grams of hydrogen}{32 grams of methanol}$

mass of hydrogen= 375 grams

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.