\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_{H_2}=0,3.2=0,6\left(g\right)\)

\(n_{CO_2}=\dfrac{28,6}{44}=0,65\left(mol\right)\)

\(n_{hh}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

x--------------->x

\(C_2H_6+\dfrac{7}{2}O_2\underrightarrow{t^o}2CO_2+3H_2O\)

y------------------>2y

\(\left\{{}\begin{matrix}x+y=0,4\\x+2y=0,65\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,25\end{matrix}\right.\)

\(V_{CH_4}=0,15.24,79=3,7185\left(l\right)\\ V_{C_2H_6}=0,25.24,79=6,1975\left(l\right)\)

Độ tan của NaCl ở 25^oC:

\(S_{NaCl}=\dfrac{m_{ct}.100\%}{m_{dm}}=\dfrac{33.100\%}{150}=22\left(g\right)\)

\(\left\{{}\begin{matrix}2p+n=50\\2p-n=10\end{matrix}\right.\)

Giải hệ pt: \(\left\{{}\begin{matrix}p=15\\n=20\end{matrix}\right.\)

Số hạt neutron trong nguyên tử Y: 20 hạt.

\(n_Z=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\) 

\(\overline{M}_Z=\dfrac{34+2}{2}=18\left(g/mol\right)\)

H₂S    34                 16

                     18

H₂      2                   16

=> \(\dfrac{n_{H_2S}}{n_{H_2}}=\dfrac{16}{16}=1\Rightarrow n_{H_2S}=n_{H_2}=\dfrac{0,15}{2}=0,075\left(mol\right)\)

\(Fe_{dư}+2HCl\rightarrow FeCl_2+H_2\)

0,075<--------------------- 0,075

=> \(n_{Fe.dư}=0,075\left(mol\right)\)

Bảo toàn S: \(n_{FeS}=n_{H_2S}=0,075\left(mol\right)\)

=> \(n_{Fe.bđ}=0,075+0,075=0,15\left(mol\right)\)

=> \(a=m_{Fe}=0,15.56=8,4\left(g\right)\)

\(n_{S.dư}=n_{S.bđ}-0,075=\dfrac{1,28}{32}=0,04\left(mol\right)\\ \Rightarrow n_{S.bđ}=0,115\left(mol\right)\Rightarrow b=m_S=0,115.32=3,68\left(g\right)\)

b) \(H\%_{Fe,S}=\dfrac{0,115.56.100\%}{0,15.56}=76,67\%\)

\(m_{H_2O_2}=\dfrac{20.3\%}{100\%}=0,6\left(g\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

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

\(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)

\(C\%_{X\left(KOH\right)}=\dfrac{0,1.56.100\%}{4,7+195,3}=2,8\%\)