Question

If 4.00 g of NaCl react with 10.00 g of AgNO3, what is the excess reactant?

Answer 1

The reaction formula of this is NaCl + AgNO3 = NaNO3 + AgCl. The mole number of NaCl is 4/58.5=0.068 mol. The mole number of AgNO3 is 10/170=0.059 mol. So the NaCl is excess.

Answer 2

Answer : The excess reagent is, $NaCl$

Explanation :

First we have to calculate the moles of $NaCl$ and $AgNO_3$.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{4g}{58.4g/mole}=0.068moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{10g}{169.9g/mole}=0.058moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NaCl+AgNO_3\rightarrow AgCl+NaNO_3$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $AgNO_3$

So, 0.068 moles of $NaCl$ react with 0.068 moles of $AgNO_3$

From this we conclude that, the moles of $AgNO_3$ are less than the NaCl. So, $AgNO_3$ is a limiting reagent because it limits the formation of products and $NaCl$ is an excess reagent.

Hence, the excess reagent is, $NaCl$