Hey
3Al(s) + 3NH4ClO4(s) ->Al2O3(s) + AlCl3(s) + 3NO(g) +6H20(g)
calculate the mass of NO and water, when 100g of aluminium is completely used up
My working out.
I found out the mole to be 1.237
so for NO
ratio is 3:3, hence 1
so
m(NO)=1.237 * (14.01 +16)
=37.12g but the answer is 111g
so for H2O
ratio is 3:6
n(H2O)= 2*1.237
=2.474 mol
m(H2O)= 2.474 * ((2*1.01)+(16))
=44.58g the answer says 133g
help thanks