$n_{H_2}=1,12/22,4=0,05mol$

$Zn+2HC\overset{t^o}{\longrightarrow}ZnCl_2+H_2$

Theo pt :

$n_{Zn}=n_{H_2}=0,05mol$

$n_{HCl}=n_{H_2}=0,01mol$

$⇒m_{Zn}=0,05.65=3,25g$

$⇒m_{HCl}=0,1.36,5=3,65g$

$2H_2+O_2\overset{t^o}{\longrightarrow}2H_2O$

Theo pt :

$n_{O_2}=1/2.n_{H_2}=1/2.0,05=0,025mol$

$⇒V_{O_2}=0,025.22,4=0,56l$

$⇒V_{kk}=5.V_{O_2}=5.0,56=2,8l$

Đáp án:

$m_{Zn}=3,25g.$

$m_{HCl}=3,65g.$

$V_{kk}=2,8l.$

Giải thích các bước giải:

$PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$  $(1)$

$n_{H_2}=\dfrac{1,12}{22,4}=0,05mol.$

$Theo$ $pt1:$ $n_{HCl}=2n_{H_2}=0,1mol.$

$⇒m_{HCl}=0,1.36,5=3,65g.$

$Theo$ $pt1:$ $n_{Zn}=n_{H_2}=0,05mol.$

$⇒m_{Zn}=0,05.65=3,25g.$

$PTPƯ:2H_2+O_2\overset{t^o}\to$ $2H_2O$  $(2)$

$Theo$ $pt2:$ $n_{O_2}=\dfrac{1}{2}n_{H_2}=0,025mol.$

$⇒V_{O_2}=0,025.22,4=0,56l.$

$⇒V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{0,56}{20\%}=2,8l.$

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