Question

Exercise #1: (4 marks) Balance each of the following redox reactions in acid condition: a) Mg(s) + H2O(g) → Mg(OH)2(s) + H2(8) b) Cr(NO3)3(aq) + Al(s) + Al(NO3)2(aq) + Cr(s) (6 marks) Exercise #2: In both cases write the balanced formula equation of the reaction and then answer to the question. a) If 25.98 mL of 0.1180 M KOH solution reacts with 52.50 mL of CH,COOH solution, what is the molarity of the acid solution? b) If 26.25 mL of 0.1850 M NaOH solution reacts with 25.00 mL of H2SO4, what is the molarity of the acid solution? Exercise # 3: (6 marks) Calculate the pressure exerted by 1.0 mol H2S behaving as (a) a perfect gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 273.15 K in 22.414 L, (ii) at 500 K in 150 mL. • Given: The van der Waals constants for HS are : a = 4.484 L .atm.mol and b=0.0434 L'mol!. Exercise # 4: (4.5 marks) A vessel of volume 22.4 L contains a mixture of 1.5 mol H2 and 2.5 mol N2 at 273.15 K. Calculate: a) the mole fractions of each component, b) The total pressure of the mixture, c) The partial pressure of each gas. Exercise #5: (4.5 marks) When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur? If so, write their balanced formula equation, total ionic equation, and net ionic equation: a) Sodium nitrate + Copper(II) sulfate → b) Ammonium bromide + Silver nitrate →

Answer 1

1- Steps followed in balancing redox reaction in acidic medium-

a- For Mg + H2O ----------> Mg(OH)2 + H2

i- Balance all the atoms other than O and H

Here already balanced

ii- Add equal number of H2O to balance the O-atom present on the other sisde

Mg + 2H2O ----------> Mg(OH)2 + H2

iii- Add equal number of H+ to balance the H-atom present on the other sisde

Already balanced

So tehe final balanced quation is-

Mg + 2H2O ----------> Mg(OH)2 + H2

b- Similarly For Cr(NO3)3 + Al ----------> Al(NO3)3 + Cr

i- Write the half cells of the reaction

Oxidation :  Al ----------> Al(NO3)3 + 3e

Reduction : Cr(NO3)3​​​​​​​ + 3e ---------->Cr

ii- Balance all the atoms other than O and H

Already balanced

iii- Add equal number of H2O to balance the O-atom present on the other sisde

Oxidation :  Al + 9H2O ----------> Al(NO3)3​​​​​​​ + 3e

Reduction : Cr(NO3)3​​​​​​​ + 3e ---------->Cr + 9H2O

iv- Add equal number of H+ to balance the H-atom present on the other sisde

Oxidation :  Al + 9H2O ----------> Al(NO3)3​​​​​​​ + 3e + 9H+

Reduction : Cr(NO3)3​​​​​​​ + 3e + 9H+ ---------->Cr + 9H2O

v- Add the two half reactions-

Oxidation :  Al + 9H2O ----------> Al(NO3)3​​​​​​​ + 3e + 9H+

Reduction : Cr(NO3)3​​​​​​​ + 3e + 9H+ ---------->Cr + 9H2O

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Overall : Cr(NO3)3 + Al ----------> Al(NO3)3 + Cr

So tehe final balanced quation is-

Cr(NO3)3 + Al ----------> Al(NO3)3 + Cr

2-

a- The chemical reaction between KOH and CH3COOH is-

KOH + CH3COOH --------------> CH3COOK + H2O

i.e 1 mole of KOH is required to neutralize 1 mole of CH3COOH

Now given molarity of KOH solution used = 0.1180 M

Volume of KOH solution used = 25.98 mL

That means moles of of KOH used = molarity * volume

= 0.1180 M * 25.98 mL

= 0.1180 mole/ 1000 mL * 25.98 mL

= 0.00307 moles

That means moles of CH3COOH neutralized = 0.00307 moles

Agai given volume of CH3COOH taken = 52.50 mL

That means in this 52.50 mL of CH3COOH solution, we have 0.00307 moles of CH3COOH preseent

So Molarity of CH3COOH = moles / volume in L

= 0.00307 moles / 52.50 mL

= 0.00307 moles / 0.05250 L

= 0.058 moles /L

= 0.058 M

b- Similarly The chemical reaction between NaOH and H2SO4 is-

2NaOH + H2SO4 --------------> Na2SO4 + 2H2O

i.e 2 mole of NaOH is required to neutralize 1 mole of H2SO4

Now given molarity of NaOH solution used = 0.1850 M

Volume of NaOH solution used = 26.25 mL

That means moles of of NaOH used = molarity * volume

= 0.185 M * 26.25mL

= 0.185 mole/ 1000 mL * 26.25 mL

= 0.0049 moles

That means moles of H2SO4 neutralized = 0.0049 moles / 2

= 0.00245‬ moles

Agai given volume of H2SO4 taken = 25 mL

That means in this 25 mL of H2SO4 solution, we have 0.00245‬ moles of H2SO4 preseent

So Molarity of H2SO4 = moles / volume in L

= 0.00245‬ moles / 25 mL

= 0.00245‬ moles / 0.025 L

= 0.098‬ moles /L

= 0.098‬ M