Question

POSTLAB QUESTIONS (Show all calculations.) 1. A student decided to try using aluminum wire instead of magnesium ribbon in this exnerima a) Write a balanced equation for the reaction. 2 Alcoy & 64ice coop) -> 200 ug cours - Bul, ca b) How much aluminum wire would the student need to produce 30.0 mL of hydrogen pas atmospheric pressure of 763 mm of Hg. (Ignore the effects of water vapor and any water colu the eudiometer for this part of the question.) c) What would the volume of the 30.0 mL of dry hydrogen gas in part b. above be if it were trapped in eudiometer tube at 22°C and an atmospheric pressure of 763 mm of Hg, and the column of water in the eudiometer was 37 mm tall? (Hint: Think about the effect of the vapor pressure of water and the height of the water column on the pressure inside the eudiometer tube.

Answer 1

1- The given reaction between solid Al and HCl solution is

2Al + 6HCl -----------> 2AlCl3 + 3H2

That means from the reaction of 2 mole Al and 6 moles HCl, we are getting 2 moles AlCl3 and 3 moles H2 gas

2- Now given the H2 gas produces has volume = 30 ml = 0.03 L

Pressure = 763 mmHg

temperature = 22^oC = 22 + 273 = 295 K

Now if we consider H2 be an Ideal gas, then we can apply the Ideal gas equation-

PV = nRT

Where P = given pressure

V = given volume

n = number of moles of gas

R = universal gas constant = 62.363 L mmHg K−1 mol−1

T = given temperature

Now putting the above values in the equation

PV = nRT

n = PV/RT

= 763 mmHg * 0.03 L / 62.363 L mmHg K−1 mol−1 * 295 K

= 0.00124 mols

That means in this process we are getting 0.00124 moles of H2

From equation 1 , we can see 3 moles of H2 is produced from = 2 moles of Al

Then 0.00124 moles of H2 is produced from = 2/3 *  0.00124

= 0.000829 moles of Al

We know mass of 1 mole Al = 27 g/mol

Then mass of  0.000829 mole Al = 0.000829 mols * 27 g/mol

= 0.0224 g of Al

That means 0.0224 g of Al wire is required to produce the Hydrogen gas.

2- Now inside an eudiometer, total pressure = P_H2O + P_H2

Here P_H2O is the pressure of H2O at 295 k = 19.8 mmHg

Again given total pressure = atmospheric pressure = 763 mmHg

Then P_H2 = total pressure - P_H2O

= 763 mmHg - 19.8 mmHg

= 743.2 mmHg‬

Now applying Daltons law of partial pressure P1V1T1 = P2V2T2

Where P1, V1, T1 are the conditions in ques-b and P2, V2, T2 are the new conditions in ques-c

Now putting the values

P1V1T1 = P2V2T2

V2 = P1V1T1 / P2T2

= 763 mmHg *30 ml * 295 k / 743.2 mmHg‬* 295 k

= 30.799 mL