NH_4ClO_4 = 14+4.1+35,5+4.16 = 18 + 35,5 + 64 = 117,5
10.27g de Al ----------- 6.117,5g[tex]NH_4ClO_4 [/tex]
x--------------- 1,325 kg
x = [tex] \frac{270.1,325}{705}= \frac{357,75}{705}= 0,507 kg[/tex]
507 g de Al.
B
[tex]Al_2O_3[/tex] = 2.27 + 3.16 = 54 + 48 = 102g
5.102 [tex]Al_2O_3[/tex] ----------- 10. 27 Al
y---------------------- 3,5.10[tex] ^{3} [/tex]
y= [tex] \frac{5.102.3,5.10^{3} }{10.27} = 6,61.10^{3} [/tex] kg