1. The problem statement, all variables and given/known data The reaction between aluminium powder and anhydrous barium nitrate is used as the propellant in some fireworks. The metal oxides and nitrogen are the only products. Which volume of nitrogen, measured under room conditions, is produced when 0.783 g of anhydrous barium nitrate reacts with an excess of aluminium? 2. Relevant equations 8Al + 6BaNO3 ---> 4Al2O3 + 6BaO + 3N2 3. The attempt at a solution Mr of BaNO3= 137.3 + 14+ (16x 3) =199.3 Moles in 0.783g = 0.783/199.3 6 moles of barium nitrate gives 3 moles of nitrogen gas so 2 moles give 1 mole of N2 gas so moles of N2 gas: 0.783/199.3*2 Volume= moles * 24000 = 47.145 cm^3 But my answer is wrong.