whn 50.0mL of .400M Ca(No3)2nis addd to 50.0mL of 0.800 M NaF, CaF2 precipitate

Question

whn 50.0mL of .400M Ca(No3)2nis addd to 50.0mL of 0.800 M NaF, CaF2 precipitate as shiwen in the net inoic down, the initial temp of both soluti is 23.0C. assuming that the reaction goes to completion and that the resulting solution has a mass of 100.00g ans a specific heat of 4.18J/gxC) calc the final temp of the solu

Ca2+(ag)+ 2 F- (s) -----> CaF2 (s)    delta H= -11.5 kJ

2.predict and baknace the folowing reaction

A. cesium acetate + aluminum nitrate ----->

B. Rubidium carbonate + hydroiodic acid ----->

Explanation / Answer

First you need to determine which reactant is the limiting one. That is, which reactant will run out first. It will determine how much heat is given off.

moles Ca2+ = M Ca2+ x L Ca2+ = (0.400)(0.0500) = 0.0200 moles Ca2+
moles F- = M F- x L F- = (0.800)(0.0500) = 0.0400 moles F-

The balanced equation tells us that it takes 2 moles of F- to react with 1 mole of Ca2+, and that's exactly what we have: 0.0400 moles F- / 0.0200 moles Ca2+ = 2/1. So both reactants will run out at the same time.

The equation also tells us that 1 mole of Ca2+ (or 2 moles of F-) will produce -11.5 kJ of heat. So how much heat will 0.0200 moles of Ca2+ produce?

0.0200 moles Ca2+ x (-11.5 kJ heat / 1 mole Ca2+) = 0.230 kJ heat = 230 J heat

This amount of heat was absorbed by the water, causing the water temperature to increase.

Heat gained by water = (mass H2O)(specific heat H2O)(Tf - Ti)
230 J = (100 g H2O)(4.18 J / g C)(Tf - 23.0)
230 = 418Tf - 9614
9844 = 418Tf
Tf = 23.55 C

Or

Moles of CaF2 formed = 0.0500 L X 0.400 mol/L = 0.0200 mol Ca2+ = 0.0200 mol CaF2

Heat released by reaction = -11.5 kJ/mol X 0.0200 mol = -0.230 kJ = -230 J

q = m c (T2-T1) = 230 J = 100 g X 4.18 J/gC X (T2 - 23.0)

T2 = 23.55 C

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