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How many grams of methanol are produced when 14 grams of carbon monoxide reacts with 1.5 grams of hydrogen gas?

User Mysterion
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1 Answer

6 votes

Answer: The mass of methanol produced is 12 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} .....(1)

  • For carbon monoxide :

Given mass of carbon monoxide = 14 g

Molar mass of carbon monoxide= 28 g/mol

Putting values in equation 1, we get:


\text{Moles of carbon monoxide}=(14g)/(28g/mol)=0.5mol

  • For hydrogen gas:

Given mass of hydrogen gas = 1.5 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:


\text{Moles of hydrogen gas}=(1.5g)/(2g/mol)=0.75mol

The chemical equation for the reaction of carbon monoxide and hydrogen gas follows:


CO+2H_2\rightarrow CH_3OH

By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of carbon monoxide

So, 0.75 moles of hydrogen gas will react with =
(1)/(2)* 0.75=0.375mol of carbon monoxide

As, given amount of carbon monoxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 1 mole of methanol

So, 0.75 moles of hydrogen gas will produce =
(1)/(2)* 0.75=0.375moles of methanol

Now, calculating the mass of methanol from equation 1, we get:

Molar mass of methanol = 32 g/mol

Moles of methanol = 0.375 moles

Putting values in equation 1, we get:


0.375mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.375mol* 32g/mol)=12g

Hence, the mass of methanol produced is 12 grams

User Jac
by
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