Question

solution stoichiometry
solve the following solutions stoichiometry problems

1. how many grams of silver chromate will precipitate when 150. ml of 0.500 m silver nitrate are added to 100. ml of 0.400 m potassium chromate?

Explanation:

Step1: Write balanced chemical equation

$$2\text{AgNO}_3(aq) + \text{K}_2\text{CrO}_4(aq) ightarrow \text{Ag}_2\text{CrO}_4(s) + 2\text{KNO}_3(aq)$$

Step2: Calculate moles of AgNO₃

$$n(\text{AgNO}_3) = M \times V = 0.500\ \text{mol/L} \times 0.150\ \text{L} = 0.0750\ \text{mol}$$

Step3: Calculate moles of K₂CrO₄

$$n(\text{K}_2\text{CrO}_4) = M \times V = 0.400\ \text{mol/L} \times 0.100\ \text{L} = 0.0400\ \text{mol}$$

Step4: Identify limiting reactant

Mole ratio: $\text{AgNO}_3:\text{K}_2\text{CrO}_4 = 2:1$
Required $\text{K}_2\text{CrO}_4$ for 0.0750 mol AgNO₃: $\frac{0.0750}{2} = 0.0375\ \text{mol} < 0.0400\ \text{mol}$
Thus, AgNO₃ is limiting.

Step5: Moles of Ag₂CrO₄ formed

$$n(\text{Ag}_2\text{CrO}_4) = \frac{1}{2}n(\text{AgNO}_3) = \frac{0.0750}{2} = 0.0375\ \text{mol}$$

Step6: Molar mass of Ag₂CrO₄

$$M(\text{Ag}_2\text{CrO}_4) = 2(107.87) + 52.00 + 4(16.00) = 331.74\ \text{g/mol}$$

Step7: Mass of Ag₂CrO₄

$$m = n \times M = 0.0375\ \text{mol} \times 331.74\ \text{g/mol} = 12.44025\ \text{g}$$

Answer:

12.4 g (rounded to 3 significant figures)