Câu 1:

$m_{S}=3,2(g) \to n_{S}=\dfrac{3,2}{32}=0,1$(mol)

Phương trình:

$S+O_2 \xrightarrow{t^0} SO_2$

Do $n_{S}=n_{O_2}=n_{SO_2}=0,1$(mol)

$\to V_{S}=V_{O_2}=V_{SO_2}=0,1.22,4=2,24(l)$

Câu 2:

$m_P=6,2(g) \to n_P=\dfrac{6,2}{31}=0,2$(mol)

Phương trình:

$4P+ 5O_2 \xrightarrow{t^0} 2P_2O_5$

$n_{O_2}=0,2.\dfrac{5}{4}=0,25$(mol)

$n_{P_2O_5}=0,2.\dfrac{2}{4}=0,1$(mol)

$\to V_{O_2}=0,25.22,4=5,6(l)$

$\to m_{P_2O_5}=0,1.142=14,2(g)$

Câu 3:

$m_{Fe}=5,6(g) \to n_{Fe}=\dfrac{5,6}{56}=0,1$(mol)

Phương trình:

$3Fe + 2O_2 \xrightarrow{t^0} Fe_3O_4$

Ta có:

$n_{O_2}=0,1.\dfrac{2}{3}=\dfrac{1}{15}$(mol)

$n_{Fe_3O_4}=0,1.\dfrac{1}{3}=\dfrac{1}{30}$(mol)

$\to V_{O_2}=\dfrac{1}{15}.22,4=1,49(l)$

$\to m_{Fe_3O_4}=\dfrac{1}{30}.232=7,73(g)$

Câu 4:

$n_{CH_4}=\dfrac{5,6}{22,4}=0,25$(mol)

Phương trình:

$CH_4 + 2O_2 \xrightarrow{t^0} CO_2 +2H_2O$

Ta có:

$n_{CH_4}=n_{CO_2}=0,25$(mol)

$\to V_{CO_2}=V_{CH_4}=0,25.22,4=5,6(l)$

$n_{H_2O}=n_{O_2}=0,25.2=0,5$(mol)

$\to V_{O_2}=0,5.22,4=11,2(l)$

$\to m_{H_2O}=18.0,5=9(g)$

Câu 5:

$m_{P_2O_5}=14,2(g)\to n_{P_2O_5}=\dfrac{14,2}{142}=0,1$(mol)

Phương trình:

$ 4P + 5O_2 \xrightarrow{t^0} 2P_2O_5$

Ta có:

$n_{O_2}=0,1.\dfrac{5}{2}=0,25$(mol)

$\to V_{O_2}=0,25.22,4=5,6$(l)

$n_{P}=0,1.\dfrac{4}{2}=0,2$(mol)

$\to m_P=31.0,2=6,2$(g)