Câu 1:
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ PbO+H_2\rightarrow\left(t^o\right)Pb+H_2O\\ 2Al+3H_2SO_{4\left(loãng\right)}\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ C_3H_8+5O_2\rightarrow\left(t^o\right)3CO_2+4H_2O\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
Bài 2:
\(a,n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2}=n_{ZnSO_4}=n_{H_2SO_4}=n_{Zn}=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ b,m_{ZnSO_4}=161.0,4=64,4\left(g\right)\\ c,n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\\ Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\\ Vì:\dfrac{0,2}{1}>\dfrac{0,4}{4}\Rightarrow Fe_3O_4dư\\ n_{Fe_3O_4\left(dư\right)}=0,2-\dfrac{0,4}{4}=0,1\left(mol\right)\\ \Rightarrow m_{Fe_3O_4\left(dư\right)}=0,1.232=23,2\left(g\right)\)
1. 2Mg + O2 \(\underrightarrow{t^o}\) 2MgO
2. \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
3. PbO + H2 \(\underrightarrow{t^o}\) H2O + Pb
4. 2Al + 3H2SO4 ( loãng ) \(\rightarrow\) Al2(SO4)3 + 3H2
5. Fe + 2HCl \(\rightarrow\) FeCl2 + H2
6. Fe2O3 + 3H2 \(\underrightarrow{t^o}\) 2Fe + 3H2O
7. C3H8 + 5O2 \(\underrightarrow{t^o}\) 3CO2 + 4H2O
8. 2KClO3 \(\underrightarrow{t^o}\) 2KCl + 3O2
