Question

I need to know the following values in the table below. All the info you need should be listed.

1. Write the balanced equation for the reaction between solid zinc and aqueous hydrochloric acid. Be sure to include physical states.

Zn(s) + 2HCl (aq)

Answer 1

1. Zn(s) + 2 HCl(aq) = ZnCl₂(aq) + H₂(g)

2.d CALCULATION OF NUMBER OF MOLES OF Zn

Number of moles of Zn = mass in g / Atomic mass in g

Trial 1: Number of moles of Zn = 0.25g/65.39g = 0.0038

Trial 2: Number of moles of Zn = 0.5g/65.39g = 0.0076

Trial 3: Number of moles of Zn = 1.00g/65.39g = 0.0152

e) CALCULATION OF NUMBER OF MOLES OF HCl

A 1M solution of HCl contains 1 mol of HCl in 1000ml of the solution

In other words,1000ml of 1M solution of HCl contains 1 mol of HCl

Trial 1 10ml of 6M solution of HCl contains 1 x10 x 6/1000 x1 = 0.06 mol of HCl

Trial 2 same as trial 1

Trial 3: 20 ml of 6M solution of HCl contains 1 x 20 x 6/1000 x1 = 0.12 mol of HCl

f) IDENTITY OF LIMITING REAGENT

THE REAGENT WITH LESSER CONCENTRATION IS THE LIMITING REAGENT

Zn(s) + 2 HCl(aq) = ZnCl₂(aq) + H₂(g)

According tothe stoichiometric equation 1mol Zn requires 2 mol HCl

In trial 1, the number of moles of Zn is 0.0038 which will require 2 x 0.0038 = 0.0076 mol HCl

In trial1, 0.06 mol HCl is available which is more than what is required.

Therfore the limiting reagent is Zn

Trial 2: Number of moles of Zn is 0.0076 which will require 2x0.0076 = 0.0152 mol HCl

In trial2 also, 0.06 mol HCl is available which is more than what is required.

Therefore the limiting reagent is Zn

Trial 3: Number of moles of Zn is 0.0152 which will require 2x0.0152 = 0.0304 mol HCl

In trial3, 0.12 mol HCl is available which is more than what is required.

Therefore the limiting reagent is Zn

g) MOLES O HYDROGEN PRODUCED

Zn(s) + 2 HCl(aq) = ZnCl₂(aq) + H₂(g)

According tothe stoichiometric equation 1mol Zn liberates1mol H₂

Trial 1: Number of moles of Zn = 0.25g/65.39g = 0.0038 ; Number of moles of H₂ = 0.0038

Trial 2: Number of moles of Zn = 0.5g/65.39g = 0.0076 ; Number of moles of H₂ = 0.0076

Trial 3: Number of moles of Zn = 1.00g/65.39g = 0.0152 ; Number of moles of H₂ = 0.0152

i) CALCULATION OF MOLAR VOLUME OF THE IDEAL HYDROGEN GAS AT ROOM TEMPERATURE

Molar volume at room temperature is the volume of 1mol of the gas at room temperature

Molar volume = Volume of the gas/ number of moles of the gas

Trial 1

Volume of the gas = 92.2 mL

Number of moles = 0.0038

Molar volume = 92.2mL/0.0038 = 24,263 mL = 24.263 L at 21.5⁰ C and 1 atmos pressure

Trial 2

Volume of the gas =184.5 mL

Number of moles = 0.0076

Molar volume = 184.5mL/0.0076 = 24,276 mL = 24.276 L at 21.5⁰ C and 1 atmos pressure

j) The molar volume of a gas is the volume of a gas at 0⁰C (273K) and 1 atmos pressure.

The volume calculated above corresponds to 21.5⁰ C and 1 atmos pressure

It should be converted into volume at 0⁰C (273K) and 1 atmos pressure.using the combined gas equation

P₀V₀/T₀ = P₁V₁/T₁

V₀ = P₁V₁ T₀/P₀T₁

Trial 1:

P₁ = 1atmos                                   P₀ = 1atmos

V₁ = 24.263 L                                V₀ = ?

T₁ = 273+ 21.5 = 294.5 K               T₁ = 273 K

V₀ = 1x 24,263x273/1x294.5 = 22.491 L

Trial 2:

V₀ = 1x 24,276 x273/1x294.5 = 22.503 L