Sodium chloride reacts with sulfuric acid. If 30.0 g of sodium chloride are available, and 14.6 g of the chloride product form, what is the percentage yield?
First, we ensure that the equation is balanced. Then, we convert the given grams to moles of reactant. We use that to determine the moles of the product from the equation. Finally, we convert those product moles back into grams.
This equation is not balanced, but because an excess of acid is given, the only pertinent fact is the ratio of sodium chloride to hydrogen chloride, which remains 1:1. The balanced reaction is:
2NaCl + H2SO4 →2HCl + N2SO4
Moles NaCl=30.0/58.5 =0.513
0.513 moles of HCl= 0.513×36.5