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Carbon monoxide ($CO$) and hydrogen (${{H}_{2}}$) react to form methanol ($C{{H}_{3}}OH$) according to the following reactions:
$CO(g)+2{{H}_{2}}(g)\to C{{H}_{3}}OH(l)$
How much $C{{H}_{3}}OH$ in mg is obtained from 0.01 mol of $CO$ and 0.08 g of ${{H}_{2}}$?

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Last updated date: 17th Jul 2024
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Answer
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Hint: A mole of a substance or particle can be defined as containing exactly $6.02214076\times {{10}^{23}}$ particles which may be atoms, molecules or ions where $6.02214076\times {{10}^{23}}$ is known as the Avogadro’s number.

Complete answer:
Mole is generally represented by the symbol mol. It is generally described as the unit of measurement for amount of substance in SI where SI stands for International System of units. It is defined on the basis of Avogadro’s number.
To calculate that how much $C{{H}_{3}}OH$ in mg is obtained from 0.01 mol of $CO$and 0.08 g of ${{H}_{2}}$, we first have to write the chemical equation given in the question i.e.
$CO(g)+2{{H}_{2}}(g)\to C{{H}_{3}}OH(l)$
Here 1 mol of $CO$and 2 mol of ${{H}_{2}}$gives 1 mol of $C{{H}_{3}}OH$
This corresponds that 0.01 mol of $CO$ and 0.02 mol of ${{H}_{2}}$ gives 0.01 mol of $C{{H}_{3}}OH$
After that if we twice the value then 0.02 mol of $CO$ and 0.04 mol of ${{H}_{2}}$ gives 0.02 mol of $C{{H}_{3}}OH$ Given values are $CO$(g) = 0.01 mol , ${{H}_{2}}$(g) = 0.08 g which is equal to 0.04 mol as mentioned above.
Thus here $CO$(g) is a limiting agent.
Therefore, $C{{H}_{3}}OH$ formed = 0.01 mol = 0.32 g or 320 mg.

Note:
Limiting reagents are defined as those substances which are completely consumed in the completion of a chemical reaction and can also be known by the name limiting agents or limiting reactants. According to the stoichiometry of chemical reactions, a fixed amount of reactants is required for the completion of the reaction.