Question

A solution of 100 mL of .200M sodium chromate is mixed with a solution of 200 mL of 0.150M strontium nitrate. What is the is the limiting reagent or reacting? How many grams of percipitate forms? What is the concentration of spectator ions in the final mixture? What is the concentration of the excess ion?

Answer 1

The balanced chemical equation

Na₂ CrO₄ (aq)+Sr(NO₃)₂(aq)--->SrCrO₄(s)+2NaNO₃(aq)

Given: 100 mL of 0.200M sodium chromate

200 mL of 0.150M strontium nitrate.

Calculate the moles of both reactants by using formula

n=C⋅V

Moles of sodium chromate = 0.200mol/L * (100/1000)L = 0.020 moles

Moles of strontium nitrate = 0.150 mol/L * (200/1000)L = 0.030 moles

Sodium chromate is limiting reagent, because it has fewer moles.

Reaction stoichiometry shows 1:1 molar relation between reactant sodium chromate and product strontium chromate

So, moles of sodium chromate reacted = moles of strontium chromate produced= 0.020 moles

Determine the mass of precipitate produced

Strontium chromate's molar mass = 203.6 g/mol

Mass of precipitate produced = moles of strontium chromate x molar mass of strontium chromate

                                                     = 0.020 moles x 203.6 g/mol = 4.072 g

4.072 grams of precipitate forms

Concentration of spectator ions in the final mixture

Na^+(aq) CrO₄^2- (aq)+Sr²⁺ (NO₃) ^1-(aq)→SrCrO₄(s) +Na⁺ (aq)NO₃^1-(aq)

Na⁺ (aq) and NO₃^1-(aq) are spectator ions, because they does not react.

Final volume of mixture = 100 ml + 200ml = 300 ml

[Na+] = 0.020 moles/ (300/1000)L = 0.067M

[NO₃^-] = 0.030 moles/(300/1000)L = 0.100 M

Concentration of the excess ion:

The 1:1 mole ratio will consume all the moles of sodium chromate and leave excess of Sr(NO₃)₂ excess of Sr(NO₃)₂ = 0.030 −0.020 = 0.010 moles

Concentration of the excess ion = excess moles/final volume

                                                          = 0.010 moles/(300/1000)L = 0.033M