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In the reaction K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 500.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?

1.5 grams

3.0 grams

480 grams

150 grams

User Bassam
by
7.5k points

1 Answer

3 votes
The balanced chemical reaction is:

K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s)

We are given the amount of the reactants to be used for the reaction. These amounts will be the starting point of our calculations.
3.0 M K2CrO4 (.500 L) = 1.5 mol K2CrO4

1.5 mol K2CrO4 (1 mol PbCrO4 / 1 mol K2CrO4) ( 323.2 g/mol)=484.8 grams PbCrO4
Therefore, the best answer from the choices is 480 grams.
User Setomidor
by
8.0k points
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