Question

3ba(no₃)₂ + 2na₃po₄ → ba₃(po₄)₂ + 6nano₃
what mass of na₃po₄ is used up as the 21.0 g ba(no₃)₂ reacts?
? g na₃po₄

Explanation:

Step1: Calculate molar mass of \( \text{Ba(NO}_3\text{)}_2 \)

Molar mass of \( \text{Ba(NO}_3\text{)}_2 \) = \( 137.33 + 2\times(14.01 + 3\times16.00) \) = \( 137.33 + 2\times(14.01 + 48.00) \) = \( 137.33 + 2\times62.01 \) = \( 137.33 + 124.02 \) = \( 261.35 \, \text{g/mol} \)
Moles of \( \text{Ba(NO}_3\text{)}_2 \) = \( \frac{\text{mass}}{\text{molar mass}} = \frac{21.0 \, \text{g}}{261.35 \, \text{g/mol}} \approx 0.08036 \, \text{mol} \)

Step2: Use stoichiometric ratio

From the reaction \( 3\text{Ba(NO}_3\text{)}_2 + 2\text{Na}_3\text{PO}_4
ightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{NaNO}_3 \), the ratio of \( \text{Ba(NO}_3\text{)}_2 \) to \( \text{Na}_3\text{PO}_4 \) is \( 3:2 \).
Let moles of \( \text{Na}_3\text{PO}_4 \) be \( n \). Then \( \frac{3}{2} = \frac{0.08036}{n} \), so \( n = \frac{2\times0.08036}{3} \approx 0.05357 \, \text{mol} \)

Step3: Calculate molar mass of \( \text{Na}_3\text{PO}_4 \)

Molar mass of \( \text{Na}_3\text{PO}_4 \) = \( 3\times22.99 + 30.97 + 4\times16.00 \) = \( 68.97 + 30.97 + 64.00 \) = \( 163.94 \, \text{g/mol} \)

Step4: Calculate mass of \( \text{Na}_3\text{PO}_4 \)

Mass = \( n \times \text{molar mass} = 0.05357 \, \text{mol} \times 163.94 \, \text{g/mol} \approx 8.78 \, \text{g} \)

Answer:

\( 8.78 \)