$n_{Zn}=19,5/65=0,3mol$

$PTHH :$

$2Zn+O_2\overset{t^o}\to 2ZnO$

a/Theo pt :

$n_{O_2}=1/2.n_{Zn}=1/2.0,3=0,15mol$

$⇒V_{O_2}=0,15.22,4=3,36l$

$V_{kk}=3,36.5=16,8l$

$b/2KClO_3\overset{t^o}\to 2KCl+3O_2$

Theo pt :

$n_{KClO_3}=2/3.n_{O_2}=2/3.0,15=0,1mol$

$⇒m_{KClO_3}=0,1.122,5=12,25g$

Đáp án:

Giải thích các bước giải:

$a,PTPƯ:2Zn+O_2\overset{t^o}\to$ $2ZnO$

$n_{Zn}=\dfrac{19,5}{65}=0,3mol.$

$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{Zn}=0,15mol.$

$⇒V_{O_2}=0,15.22,4=3,36l.$

$⇒V_{kk}=3,36:20\%=16,8l.$

$b,PTPƯ:2KClO_3\overset{t^o}\to$ $2KCl+3O_2$

$Theo$ $pt:$ $n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,1mol.$

$⇒m_{KClO_3}=0,1.122,5=12,25g.$

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