Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation: HgO + Cl2 ????HgCl2 + Cl2O What is the percent yield, if the quantity of the reactants is sufficient to produce 0.86g of Cl2O but only 0.71 g is obtained?

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Answer:

% yield =  82.5%

Explanation:

HgO + 2Cl₂ →  HgCl₂ +  Cl₂O

Our reactants are:

  • HgO and Cl₂

Our products are:

  • HgCl₂ +  Cl₂O

We do not have information about moles of reactants, but we do know the theoretical yield and the grams of product, in this case Cl₂O, we have produced.

Percent yield = (Yield produced / Theoretical yield) . 100

Theoretical yield is the mass of product which is produced by sufficent reactant. We replace data:

% yield = (0.71 g/0.86g) . 100 = 82.5%

The percent % yield =  82.5%

The balanced chemical equation will be:

HgO + 2Cl₂ →  HgCl₂ +  Cl₂O

The reactants participating in this reaction are: HgO and Cl₂

The products so formed are: HgCl₂ +  Cl₂O

[tex]\text{ Percent yield} = \frac{\text{Yield produced}}{\text{Theoretical yield}} *100[/tex]

The theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction.

[tex]\text{Percent yield}= \frac{0.71 g}{0.86g} * 100 \\\\\text{Percent yield}= 82.5\%[/tex]

Thus, the percent yield is 82.5%.

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