Đáp án:
$3/$
$a,$
$m_{O_2}=1,6g.$
$V_{O_2}=1,12l.$
$b,V_{kk}=5,6l.$
Giải thích các bước giải:
$1/$
$a,Fe_2O_3+3H_2\xrightarrow{t^o} 2Fe+3H_2O$
$b,HgO+H_2\xrightarrow{t^o} Hg+H_2O$
$c,Ag_2O+H_2\xrightarrow{t^o} 2Ag+H_2O$
$2/$
$HCl:$
$2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$H_2SO_4:$
$2Al+3H_2SO_4\xrightarrow{} 2Al_2(SO_4)_3+3H_2↑$
$Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$
$3/$
$a,PTPƯ:2Mg+O_2\xrightarrow{t^o} 2MgO$
$n_{Mg}=\dfrac{2,4}{24}=0,1mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{Mg}=0,05mol.$
$⇒m_{O_2}=0,05.32=1,6g.$
$⇒V_{O_2}=0,05.22,4=1,12l.$
$b,V_{kk}=V_{O_2}.5=1,12.5=5,6l.$
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