# Mole Sum

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1. Jan 4, 2016

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

2. Jan 4, 2016

The chemical formula for barium nitrate is $\text{Ba(NO}_3)_2$.

3. Jan 4, 2016