500 g of copper metal reacts with 2.5 L of 3.0 mol/L nitric acid solution.
The unbalanced equation for the reaction is:
Cu(s) + HNO3(aq) → H2O(g) + Cu(NO3)2(aq) + NO2(g)
Hint: Calculate how much of the copper metal remains after the reaction is complete (determine
amount consumed by the acid, then subtract that amount from the starting mass of copper (whichÂÂ
is given).ÂÂ
The Correct Answer and Explanation is :
To solve this problem, we will calculate the limiting reactant and determine how much copper reacts. Then, we calculate how much copper remains.
Step 1: Balanced Chemical Equation
The balanced equation is:
[
\text{Cu(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow 2\text{H}_2\text{O(g)} + \text{Cu(NO}_3\text{)}_2\text{(aq)} + 2\text{NO}_2\text{(g)}
]
Step 2: Calculate the moles of reactants
- Moles of copper (( \text{Cu} )):
Mass = (500 \, \text{g}), Molar mass of Cu = (63.55 \, \text{g/mol}).
[
\text{Moles of Cu} = \frac{500}{63.55} \approx 7.87 \, \text{mol}
] - Moles of nitric acid (( \text{HNO}_3 )):
Volume = (2.5 \, \text{L}), Concentration = (3.0 \, \text{mol/L}).
[
\text{Moles of HNO}_3 = 2.5 \times 3.0 = 7.5 \, \text{mol}
]
Step 3: Determine the limiting reactant
From the balanced equation, 1 mole of copper reacts with 4 moles of nitric acid.
[
\text{Moles of Cu required for 7.5 moles of HNO}_3 = \frac{7.5}{4} = 1.875 \, \text{mol}
]
Since we have (7.87 \, \text{mol Cu}), which is more than (1.875 \, \text{mol}), nitric acid is the limiting reactant.
Step 4: Calculate remaining copper
Moles of copper reacted = (1.875 \, \text{mol}).
Moles of copper remaining = (7.87 – 1.875 = 5.995 \, \text{mol}).
Mass of remaining copper:
[
\text{Mass} = 5.995 \times 63.55 \approx 381.1 \, \text{g}
]
Final Answer:
Approximately 381.1 g of copper remains.
Explanation:
Copper and nitric acid react in a 1:4 mole ratio. First, we calculated the moles of each reactant. Nitric acid was found to be the limiting reactant because it determines the extent of the reaction. Only 1.875 moles of copper react, consuming all the nitric acid. Subtracting the consumed copper from the initial amount gives the mass of unreacted copper. This approach ensures we account for the stoichiometry and conservation of mass in the reaction.