Question

How many grams of AgCl will be produced from 5.00 g of NaCl and 103 g of AgNO3? grams

Answer 1

Answer:

12.3 g.

Explanation:

Whenever a similar problem is involved, we need to write the chemical reaction representing the net change, as well as due to the fact that we need to know the stoichiometry involved.

Sodium chloride reacts with silver nitrate in a double displacement reaction to produce a cation exchange, that is, we produce silver chloride and sodium nitrate given by the following balanced chemical equation:

$AgNO_3 (aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3 (aq)$

According to the balanced chemical equation, the molar stoichiometry of silver nitrate to sodium chloride is 1 : 1.

Let's find moles of each substance to determine the limiting reactant. To do this, divide mass of each component by its molar mass:

$n_{NaCl}=\frac{5.00 g}{58.44 \frac{g}{mol} }= 0.08556 mol\\n_{AgNO_3}=\frac{103 g}{169.87 \frac{g}{mol}}= 0.6063 mol$

Since the stoichiometry of this reaction is 1 : 1, we can compare the moles directly. Notice that the number of moles of NaCl is lower than the number of moles of silver nitrate. This means NaCl is the limiting reactant.

According to the balanced chemical equation, 1 mole of NaCl produces 1 mole of AgCl, so the amount of AgCl would be:

$n_{AgCl}=n_{NaCl}=0.08556 mol$

In order to convert this into mass, let's multiply by the molar mass of AgCl:

$m_{AgCl}=0.08556 mol\cdot143.32 \frac{g}{mol}=12.3 g$