Answer to Question #121621 in General Chemistry for queen

Question

A student mixed 25.0 mL of 0.320 mol/L aqueous copper (II) sulfate, with 29.7 mL of 0.270 mol/L aqueous strontium nitrate.

a) write the balanced reaction equation for this precipitation reaction.

b) calculate the mass of precipitate that the student can expect from this reaction and identify the limiting reagent.

c) If the student collected 1.432g of precipitate, calculate the percentage yield of the reaction.

Answer 1

a) CuSO4 + Sr(NO3)2 = Cu(NO3)2 + SrSO4.

b) 0.025 x 0.32 = 0.008 mol of CuSO4

0.0297 x 0.27 = 0,008019.

0.008019-0.008 = 1.9e-5.

0.008 x (88 + 98) = 1,472 g.

Limiting reagent is CuSO4.

c) 1.432 x 100 % / 1.472 = 97,2826087 %.