- #1
Priyadarshini
- 191
- 4
Homework Statement
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?
Homework Equations
8Al + 6BaNO3 ---> 4Al2O3 + 6BaO + 3N2
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.