Question

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?

Answer 1

We need `(i)` a stoichiometric equation...........and get a yield of `77%`.

Explanation:

`CaO(s) +2HCl(aq) rarr CaCl_2(aq) + H_2O(l)`

And `(ii)` equivalent quantities of `"calcium oxide"` and `"calcium chloride"`.

`"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.