Bài 1:

$a)PTHH:2H_2O\xrightarrow{đpdd}2H_2+O_2$

$n_{H_2O}=\dfrac{7,2}{18}=0,4(mol)$

Theo PT: $n_{O_2}=\dfrac{1}{2}n_{H_2O}=0,2(mol)$

$\Rightarrow V_{O_2}=0,2.22,4=4,48(lít)$

$b)PTHH:S+O_2\xrightarrow{t^o}SO_2$

Theo PT: $n_S=n_{O_2}=0,2(mol)$

$\Rightarrow m_S=32.0,2=6,4(g)$

Bài 2:

$a)PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$

Theo PT: $n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}(mol)$

$\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232\approx 15,47(g)$

$b)$ Theo PT: $n_{O_2}=2n_{Fe_3O_4}=\dfrac{2}{15}(mol)$

$\Rightarrow V_{O_2}=\dfrac{2}{15}.22,4\approx 2,987(lít)$

$a)PTHH:2H_2O\xrightarrow{đpdd}2H_2+O_2$

$n_{H_2O}=\dfrac{7,2}{18}=0,4(mol)$

Theo PTHH $\rightarrow$ $n_{O_2}=\dfrac{1}{2}n_{H_2O}=0,2(mol)$

$\Rightarrow V_{O_2(đktc)}=0,2.22,4=4,48(lít)$ $

b)PTHH:S+O_2\xrightarrow{t^o}SO_2$ Theo PTHH

$\rightarrow$ $n_S=n_{O_2}=0,2(mol)$

$\rightarrow m_S=32.0,2=6,4(g)$

$\text{#Kudo}$