# A 1.85*g mass of calcium chloride was obtained from a 1.23*g mass of "calcium oxide" and excess hydrochloric acid. What was the percentage yield?

May 6, 2017

We need $\left(i\right)$ a stoichiometric equation...........and get a yield of 77%.

#### Explanation:

$C a O \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right)$

And $\left(i i\right)$ equivalent quantities of $\text{calcium oxide}$ and $\text{calcium chloride}$.

$\text{Yield}$ $=$ "Moles of calcium chloride"/"Moles of calcium oxide"xx100%

=((1.85*g)/(110.98*g*mol^-1))/((1.23*g)/(56.8*g*mol^-1))xx100%=77%

Sometimes when you do these equations, you might get a situation where the stoichiometry is NOT 1:1, and the yields must be appropriately calculated. This is why teachers insist that chemical reactions be stoichiometrically balanced.