QuestionMay 24, 2025

4) 5.0 g of copper metal reacts with 20.0 g of silver nitrate AgNO_(3) in solution. Calculate the t of silver formed and identify the limiting reactant. underline (1)Cu+underline (2)AgNO_(3)arrow underline (2)Ag+underline (1)Cu(NO_(3))_(2)

4) 5.0 g of copper metal reacts with 20.0 g of silver nitrate AgNO_(3) in solution. Calculate the t of silver formed and identify the limiting reactant. underline (1)Cu+underline (2)AgNO_(3)arrow underline (2)Ag+underline (1)Cu(NO_(3))_(2)
4) 5.0 g of copper metal reacts with 20.0 g of silver nitrate
AgNO_(3)
in solution. Calculate the
t of silver formed and identify the limiting reactant.
underline (1)Cu+underline (2)AgNO_(3)arrow underline (2)Ag+underline (1)Cu(NO_(3))_(2)

Solution
4.7(172 votes)

Answer

12.70 g of silver is formed; AgNO_3 is the limiting reactant. Explanation 1. Calculate moles of reactants Molar mass of Cu = 63.55 g/mol, n_{Cu} = \frac{5.0}{63.55} = 0.0787 mol. Molar mass of AgNO_3 = 169.87 g/mol, n_{AgNO_3} = \frac{20.0}{169.87} = 0.1178 mol. 2. Determine limiting reactant Reaction ratio is 1:2 (Cu:AgNO₃). Required n_{AgNO_3} for 0.0787 mol Cu = 2 \times 0.0787 = 0.1574 mol. Available n_{AgNO_3} is 0.1178 mol, so AgNO_3 is the limiting reactant. 3. Calculate moles of silver formed From the balanced equation, 2 mol AgNO_3 produces 2 mol Ag. Thus, 0.1178 mol AgNO_3 produces 0.1178 mol Ag. 4. Calculate mass of silver formed Molar mass of Ag = 107.87 g/mol. Mass of Ag = 0.1178 \times 107.87 = 12.70 g.

Explanation

1. Calculate moles of reactants<br /> Molar mass of Cu = 63.55 g/mol, $n_{Cu} = \frac{5.0}{63.55} = 0.0787$ mol. Molar mass of $AgNO_3$ = 169.87 g/mol, $n_{AgNO_3} = \frac{20.0}{169.87} = 0.1178$ mol.<br /><br />2. Determine limiting reactant<br /> Reaction ratio is 1:2 (Cu:AgNO₃). Required $n_{AgNO_3}$ for 0.0787 mol Cu = $2 \times 0.0787 = 0.1574$ mol. Available $n_{AgNO_3}$ is 0.1178 mol, so $AgNO_3$ is the limiting reactant.<br /><br />3. Calculate moles of silver formed<br /> From the balanced equation, 2 mol $AgNO_3$ produces 2 mol Ag. Thus, 0.1178 mol $AgNO_3$ produces 0.1178 mol Ag.<br /><br />4. Calculate mass of silver formed<br /> Molar mass of Ag = 107.87 g/mol. Mass of Ag = $0.1178 \times 107.87 = 12.70$ g.
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