In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver nitrate when it is mixed with an excess of sodium chloride? The equation for the reaction is below. AgNO3 + NaCl → AgCl + NaNO3

The mass of silver chloride produced is 84.37g.

The equation of the reaction is; AgNO3 + NaCl → AgCl + NaNO3

Number of moles of silver nitrate = mass of silver nitrate/Molar mass of silver nitrate

Molar mass of silver nitrate = 170 g/mol

Number of moles of silver nitrate = 100 g/170 g/mol = 0.59 moles

Mass of silver chloride =  0.59 moles × 143 g/mol = 84.37g

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The mass in grams of silver chloride, AgCl produced is 84.37 grams

Explanation:

The chemical equation for the reaction is given as follows;

AgNO₃ + NaCl   →  AgCl + NaNO₃

The mass of silver nitrate, AgNO₃, in the reaction = 100.0 g

Therefore, 1 Mole of AgNO₃ produces 1 mole of AgCl,

The molar mass of silver nitrate, AgNO₃ = 169.87 g/mol

The number of moles of silver nitrate, AgNO₃, in the reaction = 100/169.87 moles

The number of moles of silver nitrate, AgNO₃, in the reaction ≈ 0.59 moles

Since 1 mole of AgNO₃ produces 1 mole of AgCl, 0.59 moles of AgNO₃ produces 0.59 moles of AgCl

The number of moles of silver chloride, AgCl produced = 0.59 moles

The molar mass of silver chloride, AgCl = 143.32 g/mol

Therefore;

The mass of silver chloride, AgCl produced = The number of moles of silver chloride, AgCl produced × The molar mass of silver chloride, AgCl

Which gives;

The mass of silver chloride, AgCl produced = 143.32 g/mol × 0.59 moles = 84.37 g

The mass of silver chloride, AgCl produced = 84.37 g.

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