Help! Chemistry Stoichiometry Question

Question

I figured out how to balance this equation, and I think I figured out the answer to question number 1 for mass (I got 33.2 g of sodium carbonate), but how do you find the ions that are not spectator ions in the solution after the reaction? Any help is appreciated as I have spent 2 hours on this assignment.

1. Consider the unbalanced reaction between silver nitrate and sodium chromate:

AgNO₃(aq) + Na₂CrO₄(aq) → Ag₂CrO_4(s) + NaNO₃(aq)



What mass of silver chromate (in grams) will be produced when 100.0 mL of 2.00 M silver nitrate is mixed with 300.0 mL of 2.80 M sodium chromate?

Using the information from above, determine the concentration of the silver ions left in solution after the reaction is complete. 

Using the information from above, determine the concentration of the chromate ions left in solution after the reaction is complete.

Answer 1

2AgNO₃(aq) + Na₂CrO₄(aq) → Ag₂CrO_4(s) + 2NaNO₃(aq) ... balanced equation

To find mass of Ag₂CrO₄ produced, first find which reactant is LIMITING:

moles AgNO₃ present = 0.100 L x 2.00 mol/L = 0.200 moles

moles Na₂CrO₄ present = 0.300 L x 2.80 mol/L = 0.84 moles

Clearly, AgNO₃ is limiting as it takes TWO moles AgNO₃ for each ONE mole of Na₂CrO₄

mass of Ag₂CrO₄ produced = 0.200 mol AgNO₃ x 1 mol Ag₂CrO4/2mol AgNO₃ x 331.7 g/mol = 33.2 g. This agrees with your answer, except it is sodium chromate, not sodium carbonate. Congratulations.

Since the limiting reactant is AgNO₃, all of it will be used to precipitate Ag₂CrO₄ so there will be NO Ag⁺ ions left in solution.

Since the Na₂CrO₄ is NOT limiting, but is in excess, there will be some CrO₄^2- ions left in solution. Looking at the net ionic equation, we have 2Ag⁺(aq) + CrO₄^2-(aq) ==> Ag₂CrO₄(s). You started the reaction with 0.84 moles CrO₄^2- ions. Let's find the moles that reacted, and then we'll know the moles left over.

0.2 mol Ag⁺ x 1 mol CrO₄^2-/2 mol Ag⁺ = 0.1 mol CrO₄^2- used up.

moles CrO₄^2- remaining = 0.84 mol - 0.10 mol = 0.74 moles CrO₄^2- remaining.

Since the question asks for the CONCENTRATION of chromate ions, we must know divide this by the volume.

Total volume = 100.0 ml + 300.0 ml = 400.0 ml = 0.400 L

Final [CrO₄^2-] in solution = 0.74 mol/0.4 L = 1.85 M

Answer 2

The balanced equation indicates the molar ratio of the compounds to each other. So, when you balance AgNO₃(aq) + Na₂CrO₄(aq) → Ag₂CrO₄(s) + NaNO₃(aq), you find the initial ratio of 2:1 yields a final ratio of 1:2. After you calculate the number of molecules of the initial compounds from their molar masses, using the molar masses of the final compounds, you’ll be able to deduce the leftovers because the number of molecules in the final molar masses will be smaller than the initial.

I hope this helps.