Question
If 20.0 grams of copper (II) chloride CuCl2 reacts with sodium nitrate, how many moles of sodium chloride NaCl can be formed?
CuCl2 + 2 NaNO3 2 NaCl + Cu(NO3)2
CuCl2 + 2 NaNO3 2 NaCl + Cu(NO3)2
Answers
GPT-4o mini
To determine how many moles of sodium chloride (NaCl) can be formed from the reaction involving copper (II) chloride (CuCl₂) and sodium nitrate (NaNO₃), we first need to find out how many moles of CuCl₂ are present in 20.0 grams.
1. **Calculate the molar mass of CuCl₂**:
- Copper (Cu): approximately 63.55 g/mol
- Chlorine (Cl): approximately 35.45 g/mol
- Molar mass of CuCl₂ = 63.55 g/mol + 2 × 35.45 g/mol
- Molar mass of CuCl₂ = 63.55 g/mol + 70.90 g/mol = 134.45 g/mol
2. **Calculate the number of moles of CuCl₂**:
\[
\text{Number of moles of CuCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{134.45 \, \text{g/mol}} \approx 0.148 \, \text{moles}
\]
3. **Use the balanced chemical equation to find the relationship**:
The balanced reaction is:
\[
\text{CuCl₂} + 2 \, \text{NaNO₃} \rightarrow 2 \, \text{NaCl} + \text{Cu(NO₃)₂}
\]
From the equation, 1 mole of CuCl₂ produces 2 moles of NaCl.
4. **Calculate the moles of NaCl produced**:
\[
\text{Moles of NaCl} = 2 \times \text{moles of CuCl₂}
\]
\[
\text{Moles of NaCl} = 2 \times 0.148 \, \text{moles} \approx 0.296 \, \text{moles}
\]
Therefore, approximately **0.296 moles of sodium chloride (NaCl)** can be formed from the reaction of 20.0 grams of copper (II) chloride (CuCl₂) with sodium nitrate (NaNO₃).
1. **Calculate the molar mass of CuCl₂**:
- Copper (Cu): approximately 63.55 g/mol
- Chlorine (Cl): approximately 35.45 g/mol
- Molar mass of CuCl₂ = 63.55 g/mol + 2 × 35.45 g/mol
- Molar mass of CuCl₂ = 63.55 g/mol + 70.90 g/mol = 134.45 g/mol
2. **Calculate the number of moles of CuCl₂**:
\[
\text{Number of moles of CuCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{134.45 \, \text{g/mol}} \approx 0.148 \, \text{moles}
\]
3. **Use the balanced chemical equation to find the relationship**:
The balanced reaction is:
\[
\text{CuCl₂} + 2 \, \text{NaNO₃} \rightarrow 2 \, \text{NaCl} + \text{Cu(NO₃)₂}
\]
From the equation, 1 mole of CuCl₂ produces 2 moles of NaCl.
4. **Calculate the moles of NaCl produced**:
\[
\text{Moles of NaCl} = 2 \times \text{moles of CuCl₂}
\]
\[
\text{Moles of NaCl} = 2 \times 0.148 \, \text{moles} \approx 0.296 \, \text{moles}
\]
Therefore, approximately **0.296 moles of sodium chloride (NaCl)** can be formed from the reaction of 20.0 grams of copper (II) chloride (CuCl₂) with sodium nitrate (NaNO₃).