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More Last updated date: 07th Dec 2023
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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 ${{H}_{2}}$? Verified
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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.

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.