Question

BaCl2 + Na3PO4 -----> NaCl + Ba3(PO4)2.

If 0.5 moles of BaCl2 is mixed with 0.2 moles of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is?

Answer 1

Answer to the Question #47430 - Chemistry – Inorganic Chemistry

Question

$$\mathrm{BaCl_2} + \mathrm{Na_3PO_4} \longrightarrow \mathrm{NaCl} + \mathrm{Ba_3(PO_4)_2}.$$

If 0.5 moles of $\mathrm{BaCl_2}$ is mixed with 0.2 moles of $\mathrm{Na_3PO_4}$, the maximum number of moles of $\mathrm{Ba_3(PO_4)_2}$ that can be formed is?

Answer:

Balanced reaction equation is:

$$3\mathrm{BaCl_2} + 2\mathrm{Na_3PO_4} = 6\mathrm{NaCl} + \mathrm{Ba_3(PO_4)_2}$$

Make a proportion:

- 3 moles of $\mathrm{BaCl_2}$ react with 2 moles of $\mathrm{Na_3PO_4}$

- 0.5 moles of $\mathrm{BaCl_2} - x$ moles of $\mathrm{Na_3PO_4}$

$$x = \frac{0.5 \cdot 2}{3} = 0.33 \text{ moles of } \mathrm{Na_3PO_4} \text{ should react with } 0.5 \text{ moles of } \mathrm{BaCl_2}$$

We have only 0.2 moles of $\mathrm{Na_3PO_4}$, therefore it is the limiting reactant.

We need to make another proportion to calculate the maximum number of moles of $\mathrm{Ba_3(PO_4)_2}$ that can be formed by mixing 0.5 moles of $\mathrm{BaCl_2}$ with 0.2 moles of $\mathrm{Na_3PO_4}$:

- 2 moles of $\mathrm{Na_3PO_4}$ produce 1 mole of $\mathrm{Ba_3(PO_4)_2}$

- 0.2 moles of $\mathrm{Na_3PO_4} - x$ moles of $\mathrm{Ba_3(PO_4)_2}$

$$x = \frac{0.2 \cdot 1}{2} = 0.1 \text{ moles of } \mathrm{Ba_3(PO_4)_2} \text{ could be produced}$$

Answer: 0.1 moles of $\mathrm{Ba_3(PO_4)_2}$