Answer
There must be used 4.355 kg of $NH_4ClO_4$ for each kilogram of Al.
Work Step by Step
$ Al $ : 26.98 kg/kmol
$$ \frac{1 \space kmol \space Al }{ 26.98 \space kg \space Al } \space and \space \frac{ 26.98 \space kg \space Al }{1 \space kmol \space Al }$$
$ NH_4ClO_4 $ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 4 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 4 )= 117.49 kg/kmol
$$ \frac{1 \space kmol \space NH_4ClO_4 }{ 117.49 \space kg \space NH_4ClO_4 } \space and \space \frac{ 117.49 \space kg \space NH_4ClO_4 }{1 \space kmol \space NH_4ClO_4 }$$
$$ 1.000 \space kg \space Al \times \frac{1 \space kmol \space Al }{ 26.98 \space kg \space Al } \times \frac{ 3 \space kmol \space NH_4ClO_4 }{ 3 \space kmol \space Al } \times \frac{ 117.49 \space kg \space NH_4ClO_4 }{1 \space kmol \space NH_4ClO_4 } = 4.355 \space kg \space NH_4ClO_4 $$
