First write out the formula reaction: 4Al + 3O2 --- 2Al2O3 Next, find the number of moles of oxygen gas needed to oxidize 3.42 g of aluminium completely. From the cchemical equation above, 3 mol of oxygen is used to oxidize completely 4 mol of aluminium. Calculate the no. of moles of aluminium used: no. of moles of aluminium used = 3.42 g / molar mass of Al = 3.42 g / 26.98 g mol^-1 = 0.127 mol Use the mole ratio concept to find the no. of moles of oxygen required: no. of moles of oxygen required : no. of moles of aluminium used = 3 : 4 no. of moles of oxygen required / 0.127 = 3 / 4 no. of moles of oxygen required = (3 / 4)(0.127) = 0.095 mol Next, use the gas equation PV = nRT [P: Pressure of gas in Pa; V: Volume of gas in m^3; n: no. of moles; R, gas constant, 8.314 J mol^-1 K^-1; T, Temperature of gas in K] -Convert 0.975 atm to Pa: 0.975 atm = (0.975 x 101325) Pa = 98791.875 Pa -Convert 25 oC to K: 25 oC = (25 + 273) K = 298 K -Substitute into the equation PV = nRT and solve it: (98791.875)(V) = (0.095)(8.31)(298) (98791.875)(V) = 235.2561 V = 235.2561 / 98791.875 = 0.002381 m^3 = (0.002381 x 10^3) L = 2.381 L
