Question

40 POINTS PLEASE HURRY, WILL AWARD BRAINLIEST
2 QUESTIONS
1. If magnesium was the limiting reactant in this lab, calculate the theoretical yield of the gaseous product. Show all steps of your calculation. Show ALL work below. **convert g Mg from data table to moles Mg by using molar mass then use mole ratio from balanced equation
2. Determine the percent yield of this reaction, showing all steps of your calculation.

INFORMATION ABOUT QUESTION
Mass of Magnesium strip 0.028 g
Volume of gas collected 32.2 mL
Barometric pressure 1.10 atm
Room Temperature 24.0 °C
Vapor Pressure of the water 22.4 torr
Observations: none
Mg(s) + 2HCI(aq) → H2(g) + MgCI2(aq)
P= 1.07 atm
V= 0.032
n= 1.39 mol
R= 0.0821
T= 297 K

Answer 1

Answer: -

Mass of Magnesium strip 0.028 g

Atomic mass of magnesium = 24g mol-1

Number of moles of magnesium = Mass / atomic mass

= 0.028 g / 24 g mol-1

= 1.167 x 10^-3 moles

The balanced chemical equation for this process is

Mg(s) + 2HCI(aq) → H2(g) + MgCI2(aq)

We see from the equation,

1 mole of Mg will give 1 mole of H2.

1.167 x 10^-3 mol of Mg will give 1.167 x 10^-3 moles of H2.

Barometric Pressure =1.1 atm

Vapor pressure of water vapor = 22.4 torr = 0.029 atm

Pressure of dry gas P = 1.1 -0.029 =1.071 atm

Temperature T = 24 C + 273 =297 K

Using the relation

PV = nRT

V = nRT / P

= 1.167 x10^-3 mol x 0.082 L atm K−1 mol−1 x 297 K / 1.071 atm

=0.0265 L

Theoretical yield = 0.0265 L

Actual Yield = 32.2mL = 0.0322 L

% yield = actual x 100 / Theoretical

= 0.0322 x 100 / 0.0265

= 121.5%