$a)2Mg+O_2\xrightarrow{t^o}2MgO$

$b)4P+5O_2\xrightarrow{t^o}2P_2O_5$

$c)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$d)2Na+S\xrightarrow{t^o}Na_2S$

$e)2H_2O\xrightarrow{đp}2H_2+O_2$

$f)2KClO_3\xrightarrow{t^o,MnO_2}2KCl+3O_2$

$g)Cu+Cl_2\xrightarrow{t^o}CuCl_2$

$h)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$i)Mg+2HCl\to MgCl_2+H_2$

$j)2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$k)H_2+CuO\xrightarrow{t^o}Cu+H_2O$

$l)CaO+H_2O\to Ca(OH)_2$

Hóa hợp: $a,b,c,d,g,l$

Phân hủy: $e,f,h$

Thế: $i,j,k$

Đổi $1,6(tấn)=1600(kg)$

$\to n_{CaC_2}=\frac{1600.80\%}{64}=20(kmol)$

Theo sơ đồ: $n_{CH_2=CHCl}=n_{CaC_2}=20(kmol)$

$\to m_{PVC}=m_{CH_2=CHCl}=20.62,5=1250(kg)=1,25(tấn)$

Đáp án $A$

$a\big)$

$n_{Cu}=\frac{6,4}{64}=0,1(mol)$

$2Cu+O_2\xrightarrow{t^o}2CuO$

Theo PT: $n_{CuO}=n_{Cu}=0,1(mol)$

$\to m_{CuO}=0,1\times 80=8(g)$

$b\big)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{Cu}=0,05(mol)$

$\to V_{O_2(đktc)}=0,05\times 22,4=1,12(l)$

$\to V_{\rm không\,khí}=\frac{1,12}{20\%}=5,6(l)$

$D$

$2Na+2H_2O\to 2NaOH+H_2$

$Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4$

$Al(OH)_3+NaOH\to NaAlO_2+2H_2O$

(do $NaOH$ dư nên hòa tan hết kết tủa trắng $Al(OH)_3$)

$a\big)Zn+H_2SO_4\to ZnSO_4+H_2$

$b\big)$

Theo PT: $n_{H_2}=n_{Zn}=\frac{13}{65}=0,2(mol)$

$\to V_{H_2(đktc)}=0,2\times 22,4=4,48(l)$

$c\big)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

Theo PT: $n_{Fe}=\frac{2}{3}n_{H_2}=\frac{2}{15}(mol)$

$\to m_{Fe}=\frac{2}{15}\times 56\approx 7,47(g)$

$a\big)$

$n_{Fe}=\frac{16,8}{56}=0,3(mol)$

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

Theo PT: $n_{Fe_3O_4}=\frac{1}{3}n_{Fe}=0,1(mol)$

$\to m_{Fe_3O_4}=0,1.232=23,2(g)$

$b\big)$

Theo PT: $n_{O_2}=\frac{2}{3}n_{Fe}=0,2(mol)$

$\to V_{O_2}=0,2.22,4=4,48(l)$

$\to V_{kk}=4,48.5=22,4(l)$

$c\big)$

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

Theo PT: $n_{KMnO_4}=2n_{O_2}=0,4(mol)$

$\to m_{KMnO_4(dùng)}=\frac{0,4.158}{80\%}=79(g)$

$a\big)$

$ZnO+H_2\xrightarrow{t^o}Zn+H_2O$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$b\big)$

$n_{H_2}=\frac{3,92}{22,4}=0,175(2)$

Theo PT: $n_{H_2O}=n_{H_2}=0,175(mol)$

$\to y=m_{H_2O}=0,175.18=3,15(g)$

BTKL:

$m_M+m_{H_2}=m_N+m_{H_2O}$

$\to m_N=14,1+0,175.2-3,15=11,3(g)$

$\to x=11,3(g)$

Áp dụng t/d dtsbn:

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1$

$\to a=b=c=d$

$\to P=\frac{2a-a}{2a+a}+\frac{2a-a}{2a+a}+\frac{2a-a}{2a+a}+\frac{2a-a}{2a+a}=4.\frac{1}{4}=1$

Vậy $P=1$