Answer :
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 : The excess reagent is, [tex]NaCl[/tex]
Explanation :
First we have to calculate the moles of [tex]NaCl[/tex] and [tex]AgNO_3[/tex].
[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{4g}{58.4g/mole}=0.068moles[/tex]
[tex]\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{10g}{169.9g/mole}=0.058moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]NaCl+AgNO_3\rightarrow AgCl+NaNO_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]NaCl[/tex] react with 1 mole of [tex]AgNO_3[/tex]
So, 0.068 moles of [tex]NaCl[/tex] react with 0.068 moles of [tex]AgNO_3[/tex]
From this we conclude that, the moles of [tex]AgNO_3[/tex] are less than the NaCl. So, [tex]AgNO_3[/tex] is a limiting reagent because it limits the formation of products and [tex]NaCl[/tex] is an excess reagent.
Hence, the excess reagent is, [tex]NaCl[/tex]
