Question

1. An experiment shows that a 112 mL gas sample has a mass of 0.173 g at a pressure of 739 mmHg and a temperature of 30 ∘C.

What is the molar mass of the gas?

2

Aluminum oxide can be formed from its elements. 4Al(s) + 3O2(g) → 2Al2O3(s)

Part A What volume of oxygen is needed at 24.7 ∘C and 0.8200 atm  to completely react with 84.66 g of aluminum?

Answer 1

Dear friend your answers are

1. Problem answer is T = 30 + 273 = 303 K
p = 739/760=0.9723 atm
V = 0.112 L

n = pV/RT = 0.9723 x 0.112 / 0.08206 x 303 =0.004379 moles unknown gas ( PV = nRT)

molar mass = 0.173 g/ 0.004379 mol = 39.5067 g/mol

2. Convert the 84.66 g of Aluminum into moles.

moles = 84.66 g / 26.98 g/mol = 3.1378 moles

Now look at your balanced equation.

Notice that for every 4 moles of Al used you need 3 moles of O2? Another way of saying that is that for every ONE mole of Al you only need 3/4 ;that many moles of O2

So we just figured out that you have 3.1378 moles of Al to use up. So the number of moles of O2 y ou will need is 3.1378 x 3/4 which equals 2.3533 moles of O2.

Rearrange the Ideal Gas Law to this:

V = nRT / P

Substitute values into the equation:

V = [ (2.3533 mol) (0.08206 L atm mol¯¹ K¯¹) (297.7 K) ] / (0.8200 atm)

V = 70.1090 Litres

volume of oxygen is needed at 24.7 ∘C and 0.8200 atm  to completely react with 84.66 g of aluminum is 70.1090 litres.

Thank you, all the best.