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, \(NaCl\)
