$1)n_{O_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$PTHH:2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$2KClO_3\xrightarrow{t^o}2KCl+3O_2$

$2H_2O\xrightarrow{đpdd}2H_2+O_2$

Theo các PT: $n_{KMnO_4}=0,6(mol);n_{KClO_3}=0,2(mol);n_{H_2O}=0,6(mol)$

$a)m_{KMnO_4}=0,6.158=94,8(g)$

$b)m_{KClO_3}=0,2.122,5=24,5(g)$

$c)m_{H_2O}=0,6.18=10,8(g)$

$2)n_{O_2}=\dfrac{1,44}{22,4}\approx 0,06(mol)$

$PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5$

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

$C+O_2\xrightarrow{t^o}SO_2$

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

Theo PT: $n_P=0,048(mol);n_S=n_C=0,06(mol);n_{CH_4}=0,03(mol)$

$m_P=0,048.31=1,488(g);m_S=0,06.32=1,92(g)$

$m_C=0,06.12=0,72(g);m_{CH_4}=0,03.16=0,48(g)$

$3)n_{Fe}=\dfrac{16,8}{56}=0,3(mol);n_{O_2}=\dfrac{1,344}{22,4}=0,06(mol)$

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

Xét tỉ lệ: $\dfrac{n_{Fe}}{3}>\dfrac{n_{O_2}}{2}\Rightarrow Fe$ dư

Theo PT: $n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,03(mol)$

$\Rightarrow m_{Fe_3O_4}=0,03.232=6,96(g)$

$4)PTHH:2H_2O\xrightarrow{t^o}2H_2+O_2$

Theo PT: $n_{H_2}=2n_{O_2}$

Mà $n_{H_2}+n_{O_2}=\dfrac{1344}{22,4}=60(mol)$

$\Rightarrow n_{H_2}=40(mol);n_{O_2}=20(mol)$

Theo PT: $n_{H_2O}=n_{H_2}=40(mol)$

$\Rightarrow m_{H_2O}=40.18=720(g)=0,72(kg)$

$\text{Đáp án+Giải thích các bước giải:}$

$\text{Bài 1:}$

$\text{Ta có: $nO_{2}$=$\dfrac{6,72}{22,4}$= 0,3 mol}$

$\text{Câu A:}$

$\text{PTHH: $2KMnO_{4}$→$K_{2}MnO_{4}$+$MnO_{2}$+$O_{2}$}$

$\text{Theo PT ta có: $nKMnO_{4}$=2×0,3= 0,6 mol}$

$\text{⇒ $mKMnO_{4}$=0,6×158= 94,8 g}$

$\text{Câu B:}$

$\text{PTHH: $2KClO_{3}$→2KCl+$3O_{2}$}$

$\text{Theo PT ta có: $nKClO_{3}$=$\dfrac{2}{3}$×0,3= 0,2 mol}$

$\text{⇒ $mKClO_{3}$=0,2×122,5= 24,5 g}$

$\text{Câu C:}$

$\text{PTHH: $2H_{2}O$→$2H_{2}$+$O_{2}$}$

$\text{Theo PT ta có: $nH_{2}O$=2×0,3= 0,6 mol}$

$\text{⇒ $mH_{2}O$=0,6×18= 10,8 g}$

$\text{Bài 2:}$

$\text{Ta có: $nO_{2}$=$\dfrac{1,44}{22,4}$= 0,06 mol}$

$\text{PTHH: 4P+$5O_{2}$→$2P_{2}O_{5}$}$

$\text{Theo PT ta có: nP=$\dfrac{4}{5}$×0,06= 0,048 mol}$

$\text{⇒ mP=0,048×31= 1,488 g}$

$\text{PTHH: S+$O_{2}$→$SO_{2}$}$

$\text{Theo PT ta có: nS=$nO_{2}$= 0,06 mol}$

$\text{⇒ mS=0,06×32= 1,92 g}$

$\text{PTHH: C+$O_{2}$→$CO_{2}$}$

$\text{Theo PT ta có: nC=$nO_{2}$= 0,06 mol}$

$\text{⇒ mC=0,06×12= 0,72 g}$

$\text{PTHH: $CH_{4}$+$2O_{2}$→$CO_{2}$+$2H_{2}O$}$

$\text{Theo PT ta có: nC=$\dfrac{1}{2}$×0,06= 0,03 mol}$

$\text{⇒ $mCH_{4}$=0,03×16= 0,48 g}$

$\text{Bài 3:}$

$\text{PTHH: 3Fe+$2O_{2}$→$Fe_{3}O_{4}$}$

$\text{Chất thu được là Sắt(II,III) oxide}$

$\text{Ta có: nFe=$\dfrac{16,8}{56}$= 0,3 mol}$

$\text{$nO_{2}$=$\dfrac{1,344}{22,4}$= 0,06 mol}$

$\text{Lập tỉ lệ: $\dfrac{0,3}{3}$>$\dfrac{0,06}{2}$}$

$\text{⇒ Fe dư, bài toán tính theo số mol của $O_{2}$}$

$\text{Theo PT ta có: $nFe_{3}O_{4}$=$\dfrac{1}{2}$×0,06= 0,03 mol}$

$\text{⇒ $mFe_{3}O_{4}$=0,03×232= 6,96 g}$

$\text{Chúc bạn học giỏi hóa}$