Question

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

$2AgNO_3(aq) + K_2CrO_4(aq)
ightarrow Ag_2CrO_4(s) + 2KNO_3(aq)$

Step2: Calculate moles of $AgNO_3$

Moles = $M \times V$ = $0.500\ mol/L \times 0.150\ L = 0.0750\ mol$

Step3: Calculate moles of $K_2CrO_4$

Moles = $0.400\ mol/L \times 0.100\ L = 0.0400\ mol$

Step4: Identify limiting reactant

Mole ratio $AgNO_3:K_2CrO_4=2:1$. Required $K_2CrO_4$ for $AgNO_3$: $\frac{0.0750}{2}=0.0375\ mol < 0.0400\ mol$. So $AgNO_3$ is limiting.

Step5: Moles of $Ag_2CrO_4$ formed

$\frac{0.0750}{2}=0.0375\ mol$

Step6: Molar mass of $Ag_2CrO_4$

$2\times107.87 + 52.00 + 4\times16.00 = 331.74\ g/mol$

Step7: Mass of $Ag_2CrO_4$

Mass = $0.0375\ mol \times 331.74\ g/mol = 12.4\ g$

Answer:

12.4 g