An excess of zinc metal is added to 50.0 mL of a 0.100 M AgNO3 solution in a constant-pressure calorimeter. Assuming that the density of the solution is the same as for water, calculate the mass of the solution.

Answer 1

The question doesn't make sense to me. The 50 mL of the AgNO3 solution has a mass of 50.0 g assuming the density of water and the AgNO3 soln is the same; however, that is the mass before an excess of Zn metal is added. I have no idea how much the total material weighs. It all depends upon the amount of Zn added.

Answer 2

50mL of 0.1M = .005 moles of AgNO3, so based on balanced equation, that will react with .0025 moles Zn.

Calculate q(zinc) = -(q(sol) + q(cal))
q(zinc) = -(50*4.18*2.92 + 98.6*2.92)
q(zinc) = -898.19 J

then divide by .0025 moles of Zn reacting (and covert to kJ)

= -306kJ/mol

Answer 3

50mL of 0.1M = .005 moles of AgNO3, so based on balanced equation, that will react with .0025 moles Zn.

Calculate q(zinc) = -(q(sol) + q(cal))
q(zinc) = -(50*4.18*2.92 + 98.6*2.92)
q(zinc) = -898.19 J

then divide by .0025 moles of Zn reacting (and covert to kJ)

= -360kJ/mol