the balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver…

Explanation:

Step1: Calculate molar masses

The molar mass of $NaCl$: $M_{NaCl}=22.99 + 35.45=58.44\ g/mol$. The molar mass of $AgNO_3$: $M_{AgNO_3}=107.87+14.01 + 3\times16.00 = 169.88\ g/mol$.

Step2: Calculate moles of reactants

The moles of $NaCl$, $n_{NaCl}=\frac{m_{NaCl}}{M_{NaCl}}=\frac{4.00\ g}{58.44\ g/mol}\approx0.0685\ mol$. The moles of $AgNO_3$, $n_{AgNO_3}=\frac{m_{AgNO_3}}{M_{AgNO_3}}=\frac{10.00\ g}{169.88\ g/mol}\approx0.0589\ mol$.

Step3: Determine the stoichiometric ratio

From the balanced equation $NaCl + AgNO_3\rightarrow NaNO_3+AgCl$, the mole - ratio of $NaCl$ to $AgNO_3$ is $1:1$.

Step4: Identify the limiting and excess reactant

Based on the mole - ratio, for every 1 mole of $NaCl$ that reacts, 1 mole of $AgNO_3$ reacts. Since $n_{NaCl}=0.0685\ mol$ and $n_{AgNO_3}=0.0589\ mol$, $AgNO_3$ will be completely consumed first. So, $NaCl$ is the excess reactant.

Answer:

B. $NaCl$