A reaction between 2.0 g of silver nitrate and 3.0 g of sodium chloride yields how many grams of the precipitate silver chloride?

Step-by-step solution

1. 1

   Write and balance the equation first. Stoichiometry is meaningless without the mole ratio, and here the ratio is what makes the arithmetic simple:

   $\mathrm{AgNO_3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl}\downarrow + \mathrm{NaNO_3}$

   The equation is already balanced, so silver nitrate, sodium chloride and silver chloride are all in a $1:1:1$ ratio.

2. 2

   Compute the molar masses needed. Using $\mathrm{Ag}=107.87$, $\mathrm{N}=14.01$, $\mathrm{O}=16.00$, $\mathrm{Na}=22.99$, $\mathrm{Cl}=35.45\ \text{g/mol}$:

   $M(\mathrm{AgNO_3}) = 169.88, \quad M(\mathrm{NaCl}) = 58.44, \quad M(\mathrm{AgCl}) = 143.32\ \text{g/mol}$

3. 3

   Convert both reactant masses to moles. Grams cannot be compared directly — only moles can, because the balanced equation counts particles, not mass:

   $n(\mathrm{AgNO_3}) = \frac{2.0}{169.88} = 0.01177\ \text{mol}, \qquad n(\mathrm{NaCl}) = \frac{3.0}{58.44} = 0.05133\ \text{mol}$

4. 4

   Identify the limiting reactant. With a $1:1$ requirement, whichever reactant supplies fewer moles runs out first. Silver nitrate offers only $0.01177\ \text{mol}$ against $0.05133\ \text{mol}$ of sodium chloride, so $\mathrm{AgNO_3}$ is limiting and sodium chloride is in large excess — despite weighing more on the balance, because its molar mass is roughly three times smaller.

5. 5

   Convert limiting moles to moles, then grams, of product. The $1:1$ ratio carries the moles straight across, and the molar mass converts back to a mass:

   $n(\mathrm{AgCl}) = 0.01177\ \text{mol} \quad\Longrightarrow\quad m = 0.01177 \times 143.32 = 1.687\ \text{g}$

6. 6

   Round sensibly and check. The data carry two significant figures, so report $1.69\ \text{g}$ (about $1.7\ \text{g}$) of $\mathrm{AgCl}$. Mass balance confirms the result: the products are $1.69\ \text{g}$ of $\mathrm{AgCl}$ plus $1.00\ \text{g}$ of dissolved $\mathrm{NaNO_3}$, and $2.31\ \text{g}$ of sodium chloride is left over unreacted — a total of $5.0\ \text{g}$, exactly the mass put in.