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
Responses

88.75 mol
88.75 mol

6.67 mol
6.67 mol

0.3 mol
0.3 mol

322 mol
322 mol

Answer 1

To determine how many moles of sodium chloride (NaCl) can be formed from the reaction of copper(II) chloride (CuCl₂) with sodium nitrate (NaNO₃), we start by calculating how many moles of CuCl₂ are present in 20.0 grams.

1. Calculate the molar mass of CuCl₂:
- Molar mass of Cu = 63.55 g/mol
- Molar mass of Cl = 35.45 g/mol (and there are 2 Cl atoms)
- Molar mass of CuCl₂ = 63.55 g/mol + (2 x 35.45 g/mol) = 63.55 g/mol + 70.90 g/mol = 134.45 g/mol

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

3. From the balanced reaction:
\[
\text{CuCl₂} + 2 \text{NaNO₃} \rightarrow 2 \text{NaCl} + \text{Cu(NO₃)₂}
\]
We see that 1 mole of CuCl₂ produces 2 moles of NaCl.

4. Calculate the moles of NaCl produced:
\[
\text{Moles of NaCl} = 0.148 \text{ mol CuCl₂} \times 2 \, \text{mol NaCl/mol CuCl₂} \approx 0.296 \text{ mol NaCl}
\]

The closest option provided among the responses is 0.3 mol.

Thus, the answer is **0.3 mol** of sodium chloride (NaCl) can be formed.