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Sagot :
Answer:
Explanation:
There are two 250 mL aq solutions. The first solution is potassium chromate and the second is silver nitrate. The masses of the solute in each of the solutions are the same. After the reaction was completed, the precipitate was found to have a mass of 167 g. Calculate the concentration of potassium ions in the original potassium chromate solution.
K2CrO4 + AgNO3-----> Ag2CrO4 + KNO3
Ag2CrO4 HAS A MOLECULAR WEIGHT OF 2(108x2) + 52 + (4X16) =
216 + 52 + 64 =332
167gm/332= 0.51 moles which has 0.51 moles of CrO4 and required
2X0.51 moles K=1.02 moles of K
the original solution was 250 ml or 0.250L and had 1.02 moles/0.250L=
4.1 moles/L of K or 4.1 M of K
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