Answer:
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.
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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