Question

Calculate the mass in grams of the excess reagent remaining after the complete reaction of 5.67 g of Cu with 11.58 g of HNO3. Round your answer to 2 decimal places.

Answer 1

To solve this problem, we first need to determine the balanced chemical reaction between copper (Cu) and nitric acid (HNO3). The reaction generally can be represented as:

\[
3 \text{Cu} + 8 \text{HNO}_3 \rightarrow 3 \text{Cu(NO}_3\text{)}_2 + 4 \text{H}_2\text{O} + 2 \text{NO}_2
\]

From the balanced equation, we see that:

- 3 moles of Cu react with 8 moles of HNO3.

Now, let's calculate the moles of each reactant using their given masses.

### Step 1: Calculate Moles of Each Reactant

1. **Moles of Cu:**
\[
\text{Molar mass of Cu} = 63.55 \text{ g/mol}
\]
\[
\text{Moles of Cu} = \frac{5.67 \text{ g}}{63.55 \text{ g/mol}} \approx 0.0892 \text{ mol}
\]

2. **Moles of HNO3:**
\[
\text{Molar mass of HNO3} = 1.01 + 14.01 + 3 \times 16.00 = 63.01 \text{ g/mol}
\]
\[
\text{Moles of HNO3} = \frac{11.58 \text{ g}}{63.01 \text{ g/mol}} \approx 0.183 \text{ mol}
\]

### Step 2: Determine the Limiting Reagent

We can use the stoichiometry from the balanced equation to find the limiting reagent.

According to the balanced equation:
- 3 moles of Cu react with 8 moles of HNO3, which means the mole ratio is \(\frac{3 \text{ Cu}}{8 \text{ HNO3}} = \frac{0.0892}{x}\) where \(x\) is the moles of HNO3 needed.

To find out how many moles of HNO3 are required to react with 0.0892 moles of Cu, we can set up the proportion:
\[
\frac{3 \text{ Cu}}{8 \text{ HNO3}} \Rightarrow \frac{0.0892 \text{ mol Cu}}{x}
\]
Cross-multiplying gives:
\[
3x = 0.0892 \times 8 \implies x = \frac{0.0892 \times 8}{3} \approx 0.237 \text{ mol HNO3}
\]

Now we compare:
- Needed HNO3: 0.237 mol
- Available HNO3: 0.183 mol

Since \(0.183 < 0.237\), **HNO3 is the limiting reagent**.

### Step 3: Calculate Excess Reagent

From the stoichiometry of the reaction, we know that for every 8 moles of HNO3 consumed, 3 moles of Cu are also consumed. We need to calculate how many moles of Cu are required for 0.183 moles of HNO3:
\[
\text{Moles of Cu required} = \frac{3}{8} \times 0.183 \approx 0.0686 \text{ mol Cu}
\]

### Step 4: Mass of Cu Used

Now we convert moles of Cu used back to grams:
\[
\text{Mass of Cu used} = 0.0686 \text{ mol} \times 63.55 \text{ g/mol} \approx 4.36 \text{ g}
\]

### Step 5: Calculate Remaining Cu

Now we calculate the mass of Cu remaining:
\[
\text{Initial mass of Cu} = 5.67 \text{ g}
\]
\[
\text{Mass of Cu remaining} = 5.67 \text{ g} - 4.36 \text{ g} \approx 1.31 \text{ g}
\]

### Conclusion

The mass of the excess reagent (Cu) remaining after the complete reaction is:
\[
\boxed{1.31 \text{ g}}
\]