Answer: 229.8 g of Al should be used in the fuel mixture for every kilogram of [tex]NH_4ClO_4[/tex]
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
[tex]\text{Number of moles of }NH_4ClO_4=\frac{1000g}{117.5g/mol}=8.510moles[/tex]
[tex]3Al(s)+3NH_4ClO_4(s)\rightarrow Al_2O_3(s)+AlCl_3(s)+3NO(g)+6H_2O(g)[/tex]
According to stoichiometry:
3 moles of [tex]NH_4ClO_4[/tex] reacts with = 3 moles of aluminium
Thus 8.510 moles of [tex]NH_4ClO_4[/tex] react with = [tex]\frac{3}{3}\times 8.510=8.510[/tex] moles of aluminum
Mass of aluminium = [tex]moles\times {molar mass}}=8.510mol\times 27g/mol=229.8g[/tex]