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.
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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
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Calculate the number of moles of CuCl₂: \[ \text{Number of moles of CuCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 , \text{g}}{134.45 , \text{g/mol}} \approx 0.148 , \text{moles} \]
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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.
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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).
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