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Carbon monoxide (CO) and hydrogen (H2) react to form methanol (CH3OH) according to the following reactions:
CO(g)+2H2(g)CH3OH(l)
How much CH3OH in mg is obtained from 0.01 mol of CO and 0.08 g of H2?

Answer
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Hint: A mole of a substance or particle can be defined as containing exactly 6.02214076×1023 particles which may be atoms, molecules or ions where 6.02214076×1023 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 CH3OH in mg is obtained from 0.01 mol of COand 0.08 g of H2, we first have to write the chemical equation given in the question i.e.
CO(g)+2H2(g)CH3OH(l)
Here 1 mol of COand 2 mol of H2gives 1 mol of CH3OH
This corresponds that 0.01 mol of CO and 0.02 mol of H2 gives 0.01 mol of CH3OH
After that if we twice the value then 0.02 mol of CO and 0.04 mol of H2 gives 0.02 mol of CH3OH Given values are CO(g) = 0.01 mol , H2(g) = 0.08 g which is equal to 0.04 mol as mentioned above.
Thus here CO(g) is a limiting agent.
Therefore, CH3OH 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.