Find mass of a salt formed when 20 g of silver nitrate reacts with 55 g of sodium chloride.

Answer 1

The reaction occurs with silver chloride being precipitated from solution.

NaCl + AgNO₃ → AgCl ↓ + NaNO₃

First, we must make sure the reaction as written is balanced. It is balanced, and we note that one mole of NaCl reacts with one mole of AgNO₃. Once that is done, we must determine how many moles of each reactant we have before reaction.

NaCl has a molecular weight of 58.43, and 55 grams of it represents 55/58.43 moles.

So, before reaction, we have 0.94 moles of NaCl. Similarly, AgNO₃ has a molecular weight of 90.06 grams, and 20 grams of AgNO₃ represents 20/ 90.06 moles = 0.22 moles. Since NaCl and AgNO₃ react with each other on a 1:1 ratio, and since there are only 0.22 moles of AgNO₃, only 0.22 moles of NaCl are used in the reaction.

So, after reaction, what we have is:

NaCl + AgNO₃ → AgCl + NaNO₃

_Moles (0.94 - 0.22) 0.0 0.22 0.22

Both the residual NaCl and the newly formed NaNO₃ remain in solution. The amount of NaNO₃ salt formed is 0.22 moles, which equals 18.69 grams.