Answer to Question #195088 in Chemistry for Jason

Question #195088

how many grams are obtained by the reaction between 0.06 mol of silver nitrate and the corresponding amount of potassium chromate


1
Expert's answer
2021-05-19T05:40:59-0400

According to the reaction:

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

From here, number of moles of Ag2CrO4 equals:

n(Ag2CrO4) = n(AgNO3) /2 = 0.06 mol / 2 = 0.03 mol

Mass of Ag2CrO4:

m(Ag2CrO4) = n(Ag2CrO4) × Mr(Ag2CrO4) = 0.03 mol × 331.73 g/mol = 9.95 g

Number of moles of KNO3 equals:

n(KNO3) = n(AgNO3) = 0.06 mol

Mass of KNO3:

m(KNO3) = n(KNO3) × Mr(KNO3) = 0.06 mol × 101.1 g/mol = 6.07 g


Answer: 9.95 g; 6.07 g.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS