# Oxygen gas reacts with powdere

Oxygen gas reacts with powdered aluminum according to thefollowing reaction: 4Al(s)+3O2(g)→2Al2O3(s)..What volume of O2 gas(in L), measured at 795 mmHg and 16 ∘C, is required to completelyreact with 53.3 g of Al?

Oxygen gas reacts with aluminum as follow:

4Al(s) + 3O2(g)—->Al2O3(s)

Therefore, 4 mol of Al reacts with 3 mol of O2 .

no. moles of Al = mass/mol.wt              = 53.3 g/ 26.982g/mol              = 1.975 molTherefore, no. of mol of O2 = no. moles of Al x3/4                         = 1.975 mol x 3/4                         = 1.4815 mol                        Now we need to find the volume of O2 required at 795mmHg and temperature T = 16 oC

P = 795 mmHg = 795 mmHg x (1 atm/760 mmHg) = 1.046 atm ;T = 16 oC = 16+273 = 289 Kn = 1.4815 mol

From Ideal gas equation;

PV = nRTVolume V = nRT/P       = (1.4815 mol x 0.08205L.atm.mol-1·K-1 x 289 K)/1.046 atm       = 35.13 L      The volume of O2 gas required = 35.13 L

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