Question

3ba(no₃)₂ + 2na₃po₄ → ba₃(po₄)₂ + 6nano₃
21.0 g ba(no₃)₂ reacts with 10.0 g na₃po₄ according to the reaction above.
which is the limiting reactant?

1. ba(no₃)₂ 2. na₃po₄

enter the answer choice number.

Explanation:

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

Molar mass of \( \text{Ba(NO}_3\text{)}_2 \): \( \text{Ba} = 137.33 \), \( \text{N} = 14.01 \), \( \text{O} = 16.00 \).
Molar mass \( = 137.33 + 2\times(14.01 + 3\times16.00) = 261.34 \, \text{g/mol} \).
Moles \( = \frac{21.0 \, \text{g}}{261.34 \, \text{g/mol}} \approx 0.08036 \, \text{mol} \).

Step2: Calculate moles of \( \text{Na}_3\text{PO}_4 \)

Molar mass of \( \text{Na}_3\text{PO}_4 \): \( \text{Na} = 22.99 \), \( \text{P} = 30.97 \), \( \text{O} = 16.00 \).
Molar mass \( = 3\times22.99 + 30.97 + 4\times16.00 = 163.94 \, \text{g/mol} \).
Moles \( = \frac{10.0 \, \text{g}}{163.94 \, \text{g/mol}} \approx 0.0610 \, \text{mol} \).

Step3: Compare mole ratios

From the reaction: \( 3 \, \text{mol Ba(NO}_3\text{)}_2 : 2 \, \text{mol Na}_3\text{PO}_4 \).
Theoretical ratio (Ba to Na₃PO₄) \( = \frac{3}{2} = 1.5 \).
Actual ratio (Ba moles / Na₃PO₄ moles) \( = \frac{0.08036}{0.0610} \approx 1.317 \).

Since actual ratio (\( 1.317 \)) < theoretical ratio (\( 1.5 \)), \( \text{Ba(NO}_3\text{)}_2 \) is limiting (it is present in less than the required ratio relative to \( \text{Na}_3\text{PO}_4 \)).

Answer:

1