Answer:
Let's calculate it:
1. Moles of Ni(NO3)2:
\[ \text{Moles of Ni(NO}_3)_2 = \frac{0.73 \, \text{g}}{182.7 \, \text{g/mol}} \approx 0.00399 \, \text{mol} \]
2. Moles of NaNO3:
\[ \text{Moles of NaNO}_3 = 2 \times 0.00399 \, \text{mol} \approx 0.00798 \, \text{mol} \]
3. Mass of NaNO3:
\[ \text{Mass of NaNO}_3 = 0.00798 \, \text{mol} \times 85 \, \text{g/mol} \approx 0.678 \, \text{g} \]
Therefore, the mass of sodium nitrate produced is approximately 0.678 grams.
