Question

If you have 48.00g of K2CrO4, what volume of 0.1500 M BaCl2 solution is needed to completely react with all the K2CrO4?

Answer 1

Answer on Question #46599 - Chemistry - Other

Question

If you have $48.00\mathrm{g}$ of $\mathrm{K}_2\mathrm{CrO}_4$ , what volume of $0.1500\mathrm{M}$ $\mathrm{BaCl}_2$ solution is needed to completely react with all the $\mathrm{K}_2\mathrm{CrO}_4$ ?

Answer:

Balanced reaction equation is:

$$\mathrm {K} _ {2} \mathrm {C r O} _ {4} (\mathrm {s}) + \mathrm {B a C l} _ {2} (\mathrm {a q}) = \mathrm {B a C r O} _ {4} \downarrow (\mathrm {s}) + 2 \mathrm {K C l} (\mathrm {a q})$$

Molar masses of $\mathrm{K}_2\mathrm{CrO}_4$ and $\mathrm{BaCl}_2$ equal:

$$M (K _ {2} C r O _ {4}) = 2 M (K) + M (C r) + 4 M (O) = 2 \cdot 3 9. 0 9 8 + 5 2. 9 9 6 + 4 \cdot 1 5. 9 9 9 = 1 9 5. 1 8 8 \frac {g}{m o l}$$ $$M (B a C l _ {2}) = M (B a) + 2 M (C l) = 1 3 7. 3 2 7 + 2 \cdot 3 5. 4 5 = 2 0 8. 2 2 7 \frac {g}{m o l}$$

Number of moles of $\mathrm{K}_2\mathrm{CrO}_4$ equals:

$$n = \frac {m (K _ {2} C r O _ {4})}{M (K _ {2} C r O _ {4})} = \frac {4 8 . 0 0}{1 9 5 . 1 8 8} = 0. 2 4 6 m o l$$

According to reaction equation 1 mol of $\mathrm{K}_2\mathrm{CrO}_4$ reacts with 1 mol of $\mathrm{BaCl}_2$ , so 0.246 mol of $\mathrm{BaCl}_2$ is needed to completely react with all the $\mathrm{K}_2\mathrm{CrO}_4$ .

The volume of 0.1500 M $\mathrm{BaCl}_2$ solution needed is:

$$V = \frac {n (B a C l _ {2})}{C} = \frac {0 . 2 4 6}{0 . 1 5 0 0} = 1. 6 4 L$$

Answer: 1.64 L