Đáp án:

Giải thích các bước giải:

$1/$

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

$n_{O_2}=\dfrac{6,72}{22,4}=0,3mol.$

$Theo$ $pt:$ $n_{H_2O}=2n_{O_2}=0,6mol.$

$⇒m_{H_2O}=0,6.18=10,8g.$

$2/$

$PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$

$n_{Zn}=\dfrac{13}{65}=0,2mol.$

$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$

$⇒V_{H_2}=0,2.22,4=4,48l.$

chúc bạn học tốt!

1/

$n_{O_2}=6,72/22,4=0,3mol$

$PTHH : 2H_2+O_2\overset{t^o}\to 2H_2O$

Theo pt :

$n_{H_2O}=2.n_{O_2}=2.0,3=0,6mol$

$⇒m_{H_2O}=0,6.18=10,8g$

$2/n_{Zn}=13/65=0,2mol$

$PTHH :$

$Zn+2HCl\to ZnCl_2+H_2$

Theo pt :

$n_{H_2}=n_{Zn}=0,2mol$

$⇒V_{H_2}=0,2.22,4=4,48l$