Answer to Question #195088 in Chemistry for Jason

Question

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

Answer 1

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.