Câu 1:a)2Cu+O2---->2CuO b)3H2+Fe2O3---->2Fe+3H2O c)FeCl3+3NaOH---->Fe(OH)3+3NaCl d)2Al+6HCl---->2AlCl3+3H2 Câu 2:a)mAl=1,5×27=40,5g nH2O=3.10^23/6.10^23=0,5mol =>mH2O=0,5×18=9g nO2=5,6/22,4=0,25mol =>mO2=0,25×32=8g b)VCO2=1×22,4=22,4lít VH2=0,5×22,4=11,2lít nN2=7/28=0,25mol VN2=0,25×22,4=5,6lít Câu 3:a)PTHH:Zn+2HCl--->ZnCl2+H2 B)Tỉ lệ:số nguyên tử Zn:số phân tử HCl=1:2 Số phân tử HCl:số phân tử H2=1:1

Em tham khảo nha :

\(\begin{array}{l}
1)\\
2Cu + {O_2} \to 2CuO\\
3{H_2} + F{e_2}{O_3} \to 2Fe + 3{H_2}O\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
2)\\
a)\\
{m_{Al}} = 1,5 \times 27 = 40,5g\\
{n_{{H_2}O}} = \dfrac{{3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,5mol\\
{m_{{H_2}O}} = 0,5 \times 18 = 9g\\
{n_{{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{m_{{O_2}}} = 0,25 \times 32 = 8g\\
b)\\
{V_{C{O_2}}} = 1 \times 22,4 = 22,4l\\
{V_{{O_2}}} = 0,5 \times 22,4 = 11,2l\\
{n_{{N_2}}} = \dfrac{7}{{28}} = 0,25mol\\
{V_{{N_2}}} = 0,25 \times 22,4 = 5,6l\\
3)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{N_{Zn}}:{N_{HCl}} = 1:2\\
{N_{Zn}}:{N_{ZnC{l_2}}} = 1:1\\
c)\\
BTKL:\\
{m_{Zn}} + {m_{HCl}} = {m_{ZnC{l_2}}} + {m_{{H_2}}}\\
 \Rightarrow {m_{HCl}} = 13,6 + 0,2 - 6,5 = 7,3g\\
4)\\
{n_{C{O_2}}} = \dfrac{{4,4}}{{44}} = 0,1mol\\
{N_{{O_2}}} = {N_{C{O_2}}}\\
 \Rightarrow {n_{{O_2}}} = {n_{C{O_2}}} = 0,1mol\\
{V_{{O_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)