# Answer in General Chemistry for Malak #156488

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 )

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

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