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?

Question

Wyatt Atkins · Accepted Answer

WE need (i) a stoichiometric equation........and should get over #12*g# of a brick-red precipitate.

#2Ag^(+) + CrO_4^(2-) rarr Ag_2CrO_4(s)darr# Silver chromate is EXCEPTIONALLY insoluble, and will deposit with alacrity as a brick-red precipitate. And we need equivalent quantities of silver ion and chromate ion. #"Moles of silver ion"=150xx10^-3*Lxx0.500*mol*L^-1=0.075*mol# #"Moles of chromate ion"=100xx10^-3*Lxx0.400*mol*L^-1=0.040*mol# Clearly, there is EXCESS chromate ion, MORE than 1/2 equiv. And thus all the silver ion will precipitate as silver chromate. Given the stoichiometry, #0.075/2*mol# #Ag_2CrO_4# (approx. #12*g)# should precipitate...... i.e. #0.0375*molxx331.73*g*mol^-1=??*g#

Axel Alvarado · Answer

undefined Determine the limiting reactant, calculate the moles of potassium chromate and silver nitrate, use the reaction's stoichiometry to calculate the moles of formed silver chromate, and then use the molar mass of the precipitated silver chromate to convert the moles of silver chromate to grams.

Cooper Anderson · Answer

undefined To find out how many grams of silver chromate will precipitate, we first need to determine the limiting reactant between silver nitrate (AgNO3) and potassium chromate (K2CrO4). Then, we can calculate the amount of silver chromate (Ag2CrO4) formed using stoichiometry.

1. Write the balanced chemical equation for the precipitation reaction:
   $2AgNO_3 + K_2CrO_4 \rightarrow Ag_2CrO_4 + 2KNO_3$

2. Determine the moles of each reactant:
   $ \text{moles of AgNO}_3 = 0.150 \text{ L} \times 0.500 \text{ mol/L} = 0.075 \text{ mol} $
   $ \text{moles of K}_2\text{CrO}_4 = 0.100 \text{ L} \times 0.400 \text{ mol/L} = 0.040 \text{ mol} $

3. Identify the limiting reactant:
   Since $0.040 \text{ mol} < 0.075 \text{ mol}$, $K_2\text{CrO}_4$ is the limiting reactant.

4. Use the stoichiometry of the balanced equation to find the moles of silver chromate formed:
   From the balanced equation, $2 \text{ moles of AgNO}_3 : 1 \text{ mole of Ag}_2\text{CrO}_4$
   $0.040 \text{ mol} \times \frac{1 \text{ mol Ag}_2\text{CrO}_4}{2 \text{ mol AgNO}_3} = 0.020 \text{ mol Ag}_2\text{CrO}_4$

5. Calculate the mass of silver chromate formed:
   $ \text{Mass} = \text{moles} \times \text{molar mass}$
   $ \text{Mass} = 0.020 \text{ mol} \times (2 \times 107.87 \text{ g/mol} + 1 \times 51.996 \text{ g/mol}) = \text{mass of Ag}_2\text{CrO}_4$

Therefore, calculate the mass of silver chromate precipitated from the solution.