Answer:
The excess reactant is the NaCl.
Step-by-step explanation:
NaCl(aq) + AgNO3 (aq) → NaNO₃(aq) + AgCl (s) ↓
This is a precipitation reaction.
We need to convert the mass of each reactant to moles:
4 g / 58.45 g/mol = 0.068 moles of NaCl
10 g / 169.87 g/mol = 0.059 moles of nitrate.
As stoichiometry is 1 by 1, 1 mol of chloride will react with 1 mol of nitrate.
If I have 0.059 moles of nitrate, I will need the same amount of chloride and I have 0.068 moles. I still have chloride, therefore the excess reactant is the NaCl.
0.068 mol - 0.059 mol = 0.009 moles are the moles of NaCl that remains after the reaction is complete