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Calculate the mass of copper(II) oxide required when reacts with excess sulfuric acid to form \[1mol\] of \[CuS{O_4}\]. The equation for the reaction is:
\[CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\]
[Given:\[Cu = 63.5\] ;\[O = 16\] ;\[S = 32\] ]

Answer
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Hint: The stoichiometry of the reaction determines the amount of reactants used and products formed in a reaction. A mole gives the actual amount of the substance present in the reaction.

Complete step by step answer:
The reaction of copper(II) oxide with sulfuric acid forms copper sulfate and water. The corresponding reaction given is \[CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\].
Stoichiometry is an important factor which describes a quantitative relationship between the reactants and products. The stoichiometry of a reaction is dependent on the law of conservation of mass which states that total mass of reactants is equal to the total mass of products.
It is also described by the term atom economy and efficiency of a reaction. This leads to determination \[CuS{O_4}\] of the exact amount of reactant used and the exact amount of product formed.
For the given reaction the amount of product \[CuS{O_4}\] formed is given as \[1mol\]. The sulfuric acid acts as a reagent and as a solvent for this reaction which is present in excess. Thus the limiting reagent is copper (II) oxide which decides the amount of copper sulfate formed.
The reaction is a balanced chemical reaction having equal number of atoms on either side of the equation. Thus it is clear that one mole of \[CuO\] reacts with one mole of \[{H_2}S{O_4}\] and produces one mole of and one mole of water.
A mole is equal to the amount of substance which is taken in amount equivalent to the molar mass of the substance. Thus one mole of \[CuO\] is equal to the molar mass of \[CuO\]. Hence molar mass of \[CuO\] = atomic mass of \[Cu\]+ atomic mass of \[O\]= $63.5 + 16 = 79.5g$.
Hence the mass of copper(II) oxide required is \[79.5g\] .

Note: Unlike stoichiometric ratio of reagents used, some reagents are used in catalytic amount. These reagents do not take part in the reaction directly like stoichiometric reagents but catalyses the reaction faster and smoother. Knowing the stoichiometry of the reagent used in a reaction gives the amount of reactant and product.