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Determine the mass of silver chloride precipitataed when 2g of sodium chloride solution reacts with silver trioxonitrate v solution.(ag=108,n=14,o=16,na=23,cl=35.5)

Accepted Answer
Balanced chemical equation:
2NaCl(aq) + 2AgNO3(aq) → 2AgCl(s) + 2NaNO3(aq)

Moles of reactant present
Mass of NaCl = 2 g
Molar mass of NaCl = 58.44 g/mol
Moles of NaCl = mass/molar mass = 2 g/58.44 g/mol = 0.0342 mol

According to the balanced equation,
2 moles of NaCl react with 2 moles of AgNO3.
Therefore, 0.0342 mol of NaCl will react with 0.0342 mol of AgNO3.

Mass of AgCl precipitated
Molar mass of AgNO3 = 169.87 g/mol
Mass of AgNO3 = moles * molar mass = 0.0342 mol * 169.87 g/mol = 5.82 g

Using the balanced equation, we can determine the moles of AgCl precipitated:

2 moles of NaCl react with 2 moles of AgCl
Therefore, 0.0342 mol of NaCl will react with 0.0342 mol of AgCl

Molar mass of AgCl = 143.32 g/mol
Mass of AgCl precipitated = moles * molar mass = 0.0342 mol * 143.32 g/mol = 4.9 g

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