Question

Silver bromide is used to coat ordinary black-and-whitep hotographic film, while high-speed film uses silver iodide.

(a) When 96.0 mL of 5.0 g/LAgNO₃ is added to a coffee-cup calorimeter containing 96.0 mL of 5.0 g/L NaI, with both solutions at 25°C,what mass of AgI forms in gram?

(b) Use Appendix B tofind ΔH°_rxn. in kJ

(c) What is ΔT_soln (assume thevolumes are additive and the solution has the density and specific heat capacity of water)? in K

Answer 1

Answer – a) We are given, AgNO₃= 5.0 g/L , volume = 96.0 mL

For NaI = 5.0 g/L , volume = 96.0 mL

Reaction –

AgNO₃(aq)+ NaI(aq) -----> AgI(s) + NaNO₃(aq)

We need to calculate mas of AgNO₃ and NaI from the given data

For AgNO₃

1 L = 5.0 g

So, 0.096 L = ? g

= 0.48 g of AgNO₃

For NaI

1 L = 5.0 g

So, 0.096 L = ? g

= 0.48 g of NaI

Convert the mass of each to their moles

Moles of AgNO₃ = 0.48 g / 169.87 g.mol^-1

                              = 0.00283 moles

Moles of NaI = 0.48 g / 149.89 g.mol^-1­

                       = 0.00320 moles

We know the balanced equation has 1:1 mole ratio for all

So, moles of AgI = moles of AgNO₃ = 0.00283 moles

And moles of AgI = moles of NaI = 0.00320 moles

So lowest moles of AgI is 0.00283 moles, so limiting reactant is AgNO₃

And moles of AgI = 0.00283 moles

Mass of AgI = 0.00283 moles * 234.77 g/mol

               = 0.663 g of AgI

b) Now we need to find out the ΔH°_rxn

We know,

ΔH°_rxn = sum of ΔH°reactant - ΔH° product

           = [ΔH° AgI(s) + ΔH° NaNO₃(aq)] – [ΔH° AgNO₃ + ΔH° NaI(aq)]

          = [(-61.84 kJ) + (-446.2)] – [(-124.39) +(-288.03)]

          = -100.62 kJ

We know formula

ΔH°_rxn = - q = 100.62 kJ

                  = 1.0062*10⁵

q = m * C* ΔT

Mass of solution = 96+96 = 192 mL *1.00 g/mL = 192 g

So, ΔT = q / m*C

             = 1.0062*10⁵ / 192 * 4.184 J,g^-1oC.

            = 125^oC

   ΔT     = 398 K