Calculate the mass of the blood-red solid Ag 2 CrO 4 that forms when 50.0 mL of

Question

Calculate the mass of the blood-red solid Ag2CrO4 that forms when 50.0 mL of a 0.250 M silver nitrate solution is mixed with 30.0 mL of a 0.250 M potassium chromate solution.

write out the balanced chemical reaction:

calculate the number of moles of BOTH silver nitrate and potassium chromate:

determine which solution is the “limiting reactant” _________________________

based on this, how many moles of product, solid blood-red silver chromate, do you expect to form:

                                                _________________________

determine the mass of solid silver chromate you expect to form from the reaction

Explanation / Answer

The balanced equation is

2 AgNO3(aq) + K2CrO4 (aq) ---------> Ag2CrO4(s) + 2KNO3(aq)

50x0.25 30x0.25 0 0 initial mmoles

=12.5 = 7.5

the limiting reagent is silver nitrate as 7.5 mmoles of chromate requires 2x7.5 = 15 mmoles but Ag+ is only 12.5 Mmoles.

Thus AgNO3 is the limiting reagent

so 2 moles of AgNO3 gives 1 mole of Ag2CrO4 (s)

1.2 mmoles fof AgNO3 gives 6.25 mmoles of chromate ppt.

mass of silver cromate formed = 6.25 x 10-3 moles x 331.73g/mol

= 2.073 g

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