$a,PTPƯ:2Mg+O_2\xrightarrow{t^o} 2MgO$

$b,n_{Mg}=\dfrac{4,8}{24}=0,2mol.$

$Theo$ $pt:$ $n_{MgO}=n_{Mg}=0,2mol.$

$⇒m_{MgO}=0,2.40=8g.$

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

$⇒V_{O_2}=0,1.22,4=2,24l.$

$⇒V_{kk}=\dfrac{2,24}{20\%}=11,2l.$

$d,PTPƯ:2KClO_3\xrightarrow{t^o} 2KCl+3O_2$

$Theo$ $pt:$ $n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{15}mol.$

$⇒m_{KClO_3}=\dfrac{1}{15}.122,5=8,167g.$

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Đáp án:

Giải thích các bước giải:

a. PTHH : $2Mg + O_2 \xrightarrow{t^o} 2MgO$

b.

có : $n_{Mg} = \frac{4,8}{24} = 0,2(mol)$

$⇒ n_{MgO} = n_{Mg} = 0,2(mol)$
$⇒ m_{MgO} =0,2.40 = 8(gam)$
c.

Theo PT $n_{O2} = \frac{n_{Mg}}{2} = 0,1(mol)$
$⇒ V_{O2} = 0,1.22,4 = 2,24(lít)$
$⇒ V_{kk} = \frac{2,24}{20\%} = 11,2(lít)$

d.

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$⇒ n_{KClO_3} = \frac{2}{3}n_{O2} = 0,1.\frac{2}{3} = \frac{1}{15}(mol)$
$⇒ m_{KClO_3} = \frac{1}{15}.122,5 = 8,167(gam)$