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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?
A 46.8cm3
B 72.0cm3
C 93.6cm3
D 144cm3
Help !
2Al+Ba(NO3)2--->Al2O3+Bao+N2
Mole of Ba(NO3)2= 0.783/(261)=3x10^-3 mole
Mole of N2=mole of Ba(NO3)2= 3x10^-3 mole
1 mole--->24 dm^3
3x10^-3= 24x(3x10^-3)=0.072 dm^3
0.072dm^3=72 cm^3.
the answer is B. right?