Answer to Question #156488 in General Chemistry for Malak

Question

Changed the concentration of 5cm³ of HCl ( 0.5, 1, 1.5, 2 and 2.5)mol.dm^-3 and reacted

0.7 g Mg powder measured the volume of H₂ gas produced using a gas syringe for

exactly 20 seconds.

The data collected was as follows

Data Collection:

The volume of gas produced cm³ when reacting 5cm³ ( 0.5, 1, 1.5, 2, and 2.5) mol.dm^-3 of

HCl with 0.7g Mg powder

Concentration of HCl (mol.dm^-3) ------ (Average Hydrogen Produced cm³)

0.5 ------- 1.0

1.0 ------- 2.3

1.5 ------- 4.3

2.0 ------- 7.0

2.5 ------- 8.0

1. Find the theoretical yield for each concentration show only one calculation.
2. Find the actual and experimental yield for each concentration. Use 24dm3 of hydrogen gas as the experiment was conducted at RTP.
3. Find the rate of the reaction( look for the formula for the rate of the reaction and use correct units )

Answer 1

a) Mg + 2HCl --> MgCl₂ + H₂

5cm³ = 0.005dm³

= 0.005dm³

mole of Mg = mass/molar mass

= 0.7g / 24g

=0.029

M= n/v

n(mole of HCL) = 0.5moldm^-3 * 0.005dm³

n= 0.0025mol (limiting reagent)

mass = mole * molar mass of HCL

= 0.0025mol * 36.5g/mol

= 0.09125g

mole of H₂ = volume/ molar volume

1cm³ = 0.001dm³

= 0.001dm³/ 22.4dm³

= 4.464 * 10^-5mol

mass = 4.464 * 10^-5mol * 2 = 8.92 * 10^-5g

Theoretical yield 1 = 8.92 * 10^-5g / 0.09125g

= 0.00010%

Theoretical yield 2=0.0002g / 0.01825g

= 0.0101%

Theoretical yield 3 =0.0004 / 0.05325

= 0.0071%

Theoretical yield 4 =0.000625 /0.0365

= 0.0017%

Theoretical yield 5=0.0007 /0.0456

= 0.015%

Actual yield 1 = 4.464 * 10^-5mol

Actual yield 2 = 1.0 * 10^-4mol

Actual yield 3 = 2.0* 10^-4mol

Actual Yield 4 = 3.1* 10^-4mol

Actual yield 5 = 3.5 * 10^-4mol

using the first experiment

change in moles = 0.0025mol - 4.464 * 10^-5mol

= 0.0025

Rate = change in moles/ change in time

= 0.0025/ (20-0)

= 0.000012mol/s