Question

Stoichiometry Calculations: Gram to Gram Methanol, CH2OH, is an important industrial compound that is produced from the following UNBALANCED reaction/equation: CO (g) + H2 (g) → CH3OH (g). What mass of CO would be needed to produce 100 g of methanol?

a) 100 g
b) 50 g
c) 75 g
d) 200 g

Answer 1

The mass of CO needed to produce 100 g of methanol, calculated using stoichiometry and the balanced chemical equation, is approximately 87.36 g, which indicates the correct answer is closest to option C, 75 g.

Step-by-step explanation:

To calculate the mass of carbon monoxide (CO) needed to produce 100 g of methanol (CH3OH), first, we need to use the balanced chemical equation:

CO(g) + 2H₂(g) → CH3OH(g)

Now, we need to find the molar mass of methanol and carbon monoxide. The molar mass of CH3OH is approximately 32.04 g/mol, and the molar mass of CO is approximately 28.01 g/mol.

Using stoichiometry, we can convert the mass of methanol to moles and then use the stoichiometry of the reaction to find the moles of CO needed:

1. Convert 100 g of CH3OH to moles:
   100 g CH3OH × (1 mol CH3OH / 32.04 g CH3OH) = 3.12 mol CH3OH
2. Use the mole ratio from the balanced equation to find moles of CO:
   3.12 mol CH3OH × (1 mol CO / 1 mol CH3OH) = 3.12 mol CO
3. Convert moles of CO to grams of CO:
   3.12 mol CO × (28.01 g CO / 1 mol CO) = 87.36 g CO

The mass of CO needed to produce 100 g of methanol is approximately 87.36 g, which is not one of the options provided. However, the closest option to this calculated mass is option C, 75 g.