MasteringChemistry: Exam IV-Ch9-10 Adaptive Follow-Up Resources For Practice 10.
ID: 693112 • Letter: M
Question
MasteringChemistry: Exam IV-Ch9-10 Adaptive Follow-Up Resources For Practice 10.8 previous | item 12 Inext question set For Practice 10.8 Part A Calculate },n for the reaction as written. Use 100 g/mL as the density of the solution and C = 4.18 J/g·°C as the specific heat capacity. The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction: AgNO3(aq) + HCl(aq) AgCl(s) + HINo,(aq) When you combine 90.0 mL of 0.220 M AgNO3 with 90.0 mL of 0.220 M HCl in a coffee-cup calorimeter, the temperature changes from 23.30° C to 2427-C. Express your answer with the appropriate units. 45.8 mol Submit My Answers Give Up Incorrect: Try Again; 2 attempts remaining ContinuExplanation / Answer
Answer:
Given reaction is AgNO3+HCl ------>AgCl+HNO3
90 mL of 0.22M AgNO3 => moles of AgNO3=molarityx volume(L)=0.22 mol/Lx0.090 L=0.0198 mol AgNO3,
and 90 ml of 0.22 M HCl => moles of HCl=0.22 mol/Lx0.090 L=0.0198 mol HCl.
Therefore from the reaction, 1 mol of AgNO3 produces 1 mol of AgCl, so moles of AgCl=0.0198 mol.
Density of solution=1 g/L, total volume=90+90 ml=180 mL.
Then mass of solution=densityxvolume=1 g/mlx180 mL=180 g.
Specific heat C=4.18 J/g °C and temperature difference deltaT=24.27 °C-23.30 °C=0.97 °C
Therefore deltaHrxn=m x c x deltaT=(180 g)x(4.18 J/g °C)(0.97°C)=729.8 J=0.7298 kJ
Since you are producing 0.0198 moles of AgCl:
the overall Hrxn is:
Hrxn = (0.7298 kJ)/(0.0198 moles) = 36.86 kJ/mole