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If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?​

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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

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