can gap

a)
\(d_{SO_2/CH_4}=\dfrac{64}{16}=4\\ d_{SO_2/CO_2}=\dfrac{64}{44}=\dfrac{16}{11}\\ d_{SO_2/H_2}=\dfrac{64}{2}=32\\ d_{SO_2/N_2}=\dfrac{64}{28}=\dfrac{16}{7}\)

b) 

\(M_A=1,294.34=44\left(g/mol\right)\\ M_B=1,353.34=46\left(g/mol\right)\\ M_C=2.34=68\left(g/mol\right)\\ M_D=2,353.34=80\left(g/mol\right)\)

c) 

\(\%S_{\left(SO_2\right)}=\dfrac{32}{64}.100\%=50\%\\ \%S_{\left(SO_3\right)}=\dfrac{32}{80}.100\%=40\%\\ \%S_{\left(H_2SO_4\right)}=\dfrac{32}{98}.100\%=32,65\%\)

\(PTHH:4H_2+Fe_3O_4->3Fe+4H_2O\)

             0,5---->0,125------->0,375-->0,5  (mol)

\(m_{Fe_3O_4}=n\cdot M=0,125\cdot\left(56\cdot3+16\cdot4\right)=29\left(g\right)\)

\(m_{Fe}=n\cdot M=0,375\cdot56=21\left(g\right)\)

\(PTHH:2Al+6HCl-^{t^o}>2AlCl_3+3H_2\)

               0,2---->0,6-------->0,2--------->0,3     (mol)

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)

\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)

\(m_{H_2}=n\cdot M=0,3\cdot2=0,6\left(g\right)\)

\(PTHH:Zn+2HCl->ZnCl_2+H_2\)

               0,05--->0,1----->0,05----->0,05

\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

\(m_{ZnCl_2}=n\cdot M=0,05\cdot\left(65+35,5\cdot2\right)=6,8\left(g\right)\)

\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,05\cdot22,4=1,12\left(l\right)\)

mình  k hiểu lắm phần 
0,05--->0,1----->0,05----->0,05
bạn gt được không ạ

\(PTHH:CaCO_3-^{t^o}>CaO+CO_2\)

                0,5------------>0,5------->0,5 (mol)

\(n_{CaCO_3}=\dfrac{m}{M}=\dfrac{50}{40+12+16\cdot3}=0,5\left(mol\right)\)

\(n_{CaO}=0,5\cdot1:1=0,5\left(mol\right)\\ m_{CaO}=n\cdot M=0,5\cdot\left(40+16\right)=28\left(g\right)\\ n_{CO_2}=0,5:1\cdot1=0,5\left(mol\right)\\ m_{CO_2}=n\cdot M=0,5\cdot\left(12+16\cdot2\right)=22\left(g\right)\)

cho mình hỏi 
0,5 ------>0.5-------->0.5 mol
là gì vậy ạ

\(PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)

                  3-------------->3-------->4,5     (mol)

\(m_{KCl}=n\cdot M=3\cdot\left(39+35,5\right)=223,5\left(g\right)\\ V_{O_2\left(dktc\right)}=n\cdot22,4=4,5\cdot22,4=100,8\left(l\right)\)

a)
$\rm m_{O_2} = n.M = 0,25.32 = 8(g)$
$\rm m_{Na_2SO_4} = n.M = 0,3.142 = 42,6 (g)$
$\rm n_{CO_2} = \dfrac{V_{đktc}}{22,4} = \dfrac{4,48}{22,4} = 0,2 (mol)$
$\rm \Rightarrow m_{CO_2} = n.M = 0,2.44 = 8,8 (g)
$\rm n_{CO_2} = \dfrac{\text{Số phân tử}}{6.10^{23}} = \dfrac{1,8.10^{23}}{6.10^{23}} = 0,3 (mol)$
$\rm \Rightarrow m_{CO_2} = n.M = 0,3.44 = 13,2 (g)$
b)

$\rm V_{N_2(đktc)} = n.22,4 = 0,6.22,4 = 13,44 (l)$
$\rm V_{SO_2(đktc)} = n.22,4 = 0,15.22,4 = 3,36 (l)$
$\rm n_{CO_2} = \dfrac{m}{M} = \dfrac{2,8}{28} = 0,1 (mol)$
$\rm \Rightarrow V_{CO_2(đktc)} = n.22,4 = 0,1.22,4 = 2,24 (l)$
$\rm n_{CH_4} = \dfrac{\text{Số phân tử}}{6.10^{23}} = \dfrac{9.10^{23}}{6.10^{23}} = 1,5 (mol)$
$\rm \Rightarrow V_{CH_4(đktc)} = n.22,4 = 1,5.22,4 = 33,6 (l)$
c)

$\rm \text{Số nguyên tử Al} = n.6.10^{23} = 0,25.6.10^{23} = 1,5.10^{23} \text{(nguyên tử)}$
$\rm \text{Số phân tử $\rm CO_2$} = n.6.10^{23} = 0,3.6.10^{23} = 1,8.10^{23} \text{(phân tử)}$

$\rm n_{O_2} = \dfrac{6,4}{32} = 0,2 (mol)$
$\rm \Rightarrow \text{Số phân tử $\rm O_2$} = n.6.10^{23} = 0,2.6.10^{23} = 1,2.10^{23} \text{(phân tử)}$
$\rm n_{SO_2} = \dfrac{4,48}{22,4} = 0,2 (mol)$
$\rm \Rightarrow \text{Số phân tử $\rm SO_2$} = n.6.10^{23} = 0,2.6.10^{23} = 1,2.10^{23} \text{(phân tử)}$
d)
$\rm M_{Na_2SO_4} = 23.2 + 32 + 16.4 = 142 (g/mol)$
$\rm M_{Al_2(SO_4)_3} = 27.2 + (32 + 16.4).3 = 342 (g/mol)$
$\rm M_{NaOH} = 23 + 16 + 1 = 40 (g/mol)$
$\rm M_{FeCl_2} =56 + 35,5.2 = 127 (g/mol)$
$\rm M_{Al_2O_3} = 27.2 + 16.3 = 102 (g/mol)$
e)

$\rm n_{Al} = \dfrac{m}{M} = \dfrac{5,4}{27} = 0,2 (mol)$
$\rm n_{Ba(NO_3)_2} = \dfrac{m}{M} = \dfrac{26,1}{261} = 0,1 (mol)$
$\rm n_{CO_2} = \dfrac{V_{(đktc)}}{22,4} = \dfrac{13,44}{22,4} = 0,6 (mol)$

\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)

áp dụng ĐLBTKL ta có

\(m_P+m_{O_2}=m_{P_2O_5}\\ =>m_{O_2}=m_{P_2O_5}-m_P\\ =>m_{O_2}=12,7-3,1\\ =>m_{O_2}=9,6\left(g\right)\)