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$Fe+2HCl\rightarrow FeCl_2+H_2$
$0,4$ $0,8$ $0,4$ $0,4$
$m_{HCl}=\frac{200\times14,6}{100}=29,2(g)$
$n_{HCl}=\frac{29,2}{36,5}=0,8(mol)$
$m_{Fe}=0,4\times56=22,4(g)$
$V_{H_2}=0,4\times22,4=8,96(lit)$
$m_{ddspu}=m_{Fe}+m_{dd HCl}-m_{H_2}=22,4+200-0,4\times2=221,6(g)$
$C$%$_{FeCl_2} =\frac{0,4\times127\times100}{221,6}\approx 22,92$%
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Trả lời 07-05-14 12:11 PM
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