Question

Oxygen gas reacts with powdered aluminum according to the reaction:
4Al(s)+3O2(g)→2Al2O3(s)

What volume of O2 gas (in L), measured at 774  Torr and 29  ∘C , completely reacts with 52.0  g of Al? (answer in Litres)

Answer 1

3 moles of oxygen gas reacts with 4 moles of aluminum.

Moles of oxygen gas that reacts with 1 mole of Al = 4/3

= 1.33 moles of oxygen

Number of moles = mass/molar mass

Molar mass of Al = 27 g/mol

Moles of Al = 52/27

= 1.93 moles Al

Moles of oxygen required = 1.93 × 1.33

= 2.57 moles oxygen.

According to ideal gas equation, PV = nRT PaPressure V - Volume n-Moles

To absolute temperature R> Gas Constant (0.08206 Latm K Imot')

P = 774 Tour V=? T = 29°C n=2.57 = 29+273K latm = 760 tour 302K P=1.02 atm. PV=nRT 1.02 X V = 2.57X 0.08206x302 V= 2.57X 0.08206X 302 1.02 [V=62.4L