When 10.0 g of NaCl is added to an aqueous solution of AgNO₃, 24.5 g AgCl will be precipitated.
The balanced chemical reaction is:
NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)
Molar mass of NaCl = 58.5 g
Molar mass of AgCl = 143.5 g
Number of moles of NaCl = Given Mass
Molecular Mass
= 10.0 g ÷ 58.44 g/mol
= 0.171 moles
From the reaction,
58.5 g of NaCl produces 143.5 g of AgCl
10.0 g of NaCl will produce 143.5 × 10.0 = 24.5 g AgCl
58.5
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