Đáp án + Giải thích các bước giải:

1)

a) $n_{Fe} = \dfrac{m_{Fe}}{M_{Fe}} = \dfrac{28}{56} = 0,5\text{ mol}$
$n_{Cu} = \dfrac{m_{Cu}}{M_{Cu}} = \dfrac{64}{64} = 1\text{ mol}$
$n_{Al} = \dfrac{m_{Al}}{M_{Al}} = \dfrac{5,4}{27} = 0,2\text{ mol}$

b) $m_{O_2} = n_{O_2} . M_{O_2} = 3 . 2 . 16 = 96g$

$m_{CuSO_4} = n_{CuSO_4} . M_{CuSO_4} = 0,5 . (64 + 32 + 4 . 16) = 0,5 . 160 = 80g$

c) $V_{CO_2} = n_{CO_2} . 22,4 = 0,175 . 22,4 = 3,92l$
$V_{H_2} = n_{H_2} . 22,4 = 1,25 . 22,4 = 28l$ 

1) $n_{Fe}=\frac{m}{M} = \frac{13}{28}mol$

$n_{Cu}\frac{m}{M} =0,1mol$

$n_{Al}=\frac{m}{M} =0,2mol$

b, $m_{O_2}=n.M=96g$

$m_{CuSO_4}=n.M=80g$

c,

$V_{CO_2}=n.22,4=3,92l$

$V_{H_2}=n.22,4=28l$