The question is incomplete, the complete question is;
AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq)
1- If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?
Answer:
1.8 g
Explanation:
Given the equation;
AgNO3(aq)+NaCl(aq)→AgCl(s)↓+NaNO3(aq)
We can see that the reaction is 1:1 Hence, 1 mole of sodium chloride yielded 1 mole of the precipitate(AgCl).
If this is so,
Number of moles of precipitate formed = 4.52g/143.32 g/mol
Number of moles of precipitate formed = 0.0315 moles
Hence, 0.0315 moles of precipitate was formed by 0.0315 moles of NaCl
Therefore;
Mass of NaCl reacted = 0.0315 moles * 58.5 g/mol = 1.8 g
