Question

What volume of O2 gas (in L), measured at 781 mmHg and 33 ∘C, is required to completely react with 51.0 g of Al?

Answer 1

The balanced equation is

4 Al + 3 O₂ -----> 2 Al₂O₃

Number of moles of Al = 51.0 g / 27.0 g/mol = 1.89 mol

From the balanced equation we can say that

4 mole of Al requires 3 mole of O2 so

1.89 mole of Al will require

= 1.89 mole of Al *(3 mole of O2 / 4 mole of Al)

= 1.42 mole of O2

PV = nRT

where, P = 781 mmHg = 1.03 atm

V = volume = ?

n = number of moles of O2 = 1.42 mol

R = Gas constant

T = temperature = 33 + 273 = 306 K

1.03 * V = 1.42 * 0.0821 * 306

1.03 * V = 35.7

V = 35.7 / 1.03 = 34.7 L

Therefore, the volume of O2 required would be 34.7 L