Question

For the reaction, $CaO + 2HCl \to CaC{l_2} + {H_2}O$
$1.23{\text{g}}$ of $CaO$ is reacted with excess hydrochloric acid and $1.85{\text{g}}$ of $CaC{l_2}$ is formed. What is percent yield?

Answer 1

To solve this question, you must recall basic stoichiometric fundamentals. Stoichiometry is based on the law of conservation of mass which suggests that the total mass of reactants is equal to the total mass of products. Thus, if the amount of the separate reactants are known to us, then the amount of products can be determined.
Formula used: ${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
{\text{% yield = }}\dfrac{{{\text{actual mass}}}}{{{\text{calculated mass}}}}{{ \times 100}}

Complete step by step answer:
In the given reaction, $CaO + 2HCl \to CaC{l_2} + {H_2}O$.
One mole of calcium oxide gives one mole of calcium chloride.
In the question, it is given that $1.23{\text{g}}$ of $CaO$ gives $1.85{\text{g}}$ of $CaC{l_2}$ (actual yield)
To calculate the percent yield, we must find the theoretical yield of calcium chloride in the reaction.
Molar mass of $CaO = 56{\text{g}}$
Molar mass of $CaC{l_2} = 111{\text{g}}$
Since, $56{\text{g}}$of $CaO$ gives $111{\text{g}}$ of $CaC{l_2}$
Thus, we can find:
$1.23{\text{g}}$ of $CaO$ produces $CaC{l_2}$ $= \dfrac{{111}}{{56}} \times 1.23 = 2.43{\text{g}}$
Thus, the theoretical yield$= 2.43{\text{g}}$
And the actual yield$= 1.85{\text{g}}$
Therefore, the percent yield is given by:
$\% {\text{yield}} = \dfrac{{1.85}}{{2.43}} \times 100$
$\therefore \% {\text{yield}} = 76.1$.

Note:
Stoichiometry is based upon the very basic laws of chemistry that help to understand it better, namely, the law of conservation of mass, the law of definite proportions (the law of constant composition), the law of reciprocal proportions and the law of multiple proportions . In general, chemicals combine in definite ratios in the chemical reactions. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, thus the amount of each element must be the same throughout the entire reaction. For instance, the number of atoms of a given element X on the reactant side must be equal to the number of atoms of that element on the product side, irrespective of whether or not all of those atoms are involved in a reaction.