\(M_A=23.2=46g/mol\\ n_{CO_2}=\dfrac{8,8}{44}=0,2mol\Rightarrow n_C=n_{CO_2}=0,2mol\\ \Rightarrow n_O=2n_{CO_2}=0,4mol\\ n_{H_2O}=\dfrac{5,4}{18}=0,3mol\\ \Rightarrow n_H=2n_{H_2}=0,6mol\\ \Rightarrow n_O=n_{H_2O}=0,3mol\\ n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\\ BTKL:m_A+m_{O_2}=m_{CO_2}+m_{H_2O}\\ \Rightarrow m_A=m_{H_2O}+m_{CO_2}-m_{O_2}\\ =8,8+5,4-0,3.32=4,6g\\ Có:m_{O_2}< m_{O\left(H_2O\right)}+m_{O\left(CO_2\right)}\\ \left(0,3.32< 0,4.16+0,3.16\right)\\ \Rightarrow A.có.O\\ m_{O\left(A\right)}=4,6-0,2.12-0,6.1=1,6g\\ CTHH\left(A\right):C_xH_yO_z\)
Ta có tỉ lệ:

\(\dfrac{0,2.12}{12x}=\dfrac{0,6.1}{y}=\dfrac{1,6}{16z}=\dfrac{4,6}{46}\\ \Rightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=1\end{matrix}\right.\)

Vậy CTHH A cần tìm là \(C_2H_6O\)

\(a)n_{Zn}=a;n_{Fe}=b\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ a.....2a.......a........a\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.....2b.......b........b\\ \Rightarrow\left\{{}\begin{matrix}65a+56b=24,2\\136a+127b=52,6\end{matrix}\right.\\ \Rightarrow a=b=0,2mol\\ n_{H_2}=0,2+0,2=0,4mol\\ V=V_{H_2}=0,4.22,4=8,96l\\ n_{HCl}=0,2.2+0,2.2=0,8mol\\ 200ml=0,2l\\ C_{M_{HCl}}=\dfrac{0,8}{0,2}=4M\\ b)n_{CuO}=x;n_{Fe_2O_3}=y\\ CuO+H_2\xrightarrow[]{t^0}Cu+H_2O\\ x.....x.......x.....x\\ Fe_2O_3+3H_2\xrightarrow[]{t^0}2Fe+3H_2O\\ y.........3y......2y.....3y\)
\(\Rightarrow\left\{{}\begin{matrix}x+3y=0,4\\80x=160y\end{matrix}\right.\\ \Rightarrow x=0,16mol;y=0,08mol\\ m=m_{CuO}+m_{Fe_2O_3}=0,16.80+0,08.160=25,6g\)

\(16,8l\left(đktc\right)?\\ n_{KK}=\dfrac{16,8}{22,4}=0,75mol\\ n_{O_2}=0,75\cdot20\%:100\%=0,15mol\\ m_{O_2}=0,15.32=4,8g\\ BTKL:m_X+m_{O_2}=m_Y\\ \Rightarrow m_Y=m_X+m_{O_2}\\ \Rightarrow m_Y=m=11,2+4,8=16g\)

\(n_{H_2}=\dfrac{V}{24,79}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Gọi kim loại cần tìm là M;

\(M+2HCl=MCl_2+H_2\)

Vậy cần 0,15 mol M để có được 0,15 mol \(H_2\)

\(M_M=\dfrac{m_M}{n_M}=\dfrac{3,65}{0,15}\approx24\)

Vậy tên kim loại là Magnesium

\(a.CTHH\left(muối.ngậm.nước\right):FeSO_4.nH_2O\\ n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,05mol\\ n_{muối.ngâm.nước}=n_{FeSO_4}=0,05mol\\ M_{muối.ngậm.nước}=\dfrac{13,9}{0,05}=278g/mol\\ M_{FeSO_4.nH_2O}=152+18n=278g/mol\\ \Rightarrow n=7\\ \Rightarrow CTHH\left(muối.ngậm.nước\right):FeSO_4.7H_2O\\ b.200ml=0,2l\\ C_{M_{H_2SO_4}}=\dfrac{0,05}{0,2}=0,25M\)

giúp mình với ạ

a) \(n_{NaOH}=C_{M,NaOH}\cdot V_{NaOH}=2\cdot0,2=0,4\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\) 

Theo PTHH: \(n_{HCl}=n_{NaOH}=0,4\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{n_{HCl}}{C_{M,HCl}}=\dfrac{0,4}{1,2}=\dfrac{1}{3}\left(l\right)\)  

b) Thể tích dung dịch sau phản ứng là:

\(0,2+\dfrac{1}{3}=\dfrac{8}{15}\left(l\right)\)

Theo PTHH: \(n_{NaCl}=n_{NaOH}=0,4\left(mol\right)\) 

\(\Rightarrow C_{M,NaOH}=\dfrac{0,4}{\dfrac{8}{15}}=0,75\left(M\right)\) 

dòng cuối là NaCl :)))

\(n_{P_2O_5}=\dfrac{22,72}{142}=0,16mol\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\\ a.n_P=0,16\cdot\dfrac{4}{2}=0,32mol\\ m=m_P=0,32.31=9,92g\\ b.n_{O_2}=0,16\cdot\dfrac{5}{2}=0,4mol\\ 0V=V_{O_2}=0,4.22,4=8,96l\)

5,6 tấn = 5600 (kg)

Ta có: \(n_{Fe}=\dfrac{5600}{56}=100\left(kmol\right)\)

PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

Theo PT: \(n_{Fe_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Fe}=50\left(kmol\right)\)

Mà: H = 90%

\(\Rightarrow n_{Fe_2O_3\left(TT\right)}=\dfrac{50}{90\%}=\dfrac{500}{9}\left(kmol\right)\)

\(\Rightarrow m_{Fe_2O_3\left(TT\right)}=\dfrac{500}{9}.160=\dfrac{80000}{9}\left(kg\right)\)

⇒ m _quặng = \(\dfrac{m_{Fe_2O_3\left(TT\right)}}{30\%}\approx29629,6\left(kg\right)\) ≃ 29,63 (tấn)

Ta có: m_H2O = 0,18 (g) 

\(\Rightarrow n_{H_2O}=\dfrac{0,18}{18}=0,01\left(mol\right)\Rightarrow n_H=0,01.2=0,02\left(mol\right)\)

m_CO2 = 0,44 (g)

\(\Rightarrow n_{CO_2}=\dfrac{0,44}{44}=0,01\left(mol\right)=n_C\)

Có: \(n_A=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

Gọi CTPT của A là C_xH_y.

\(x=\dfrac{0,01}{0,005}=2\)

\(y=\dfrac{0,02}{0,005}=4\)

Vậy: A là C₂H₄.

\(39,6g?\\ CTPT\left(A\right):C_xH_y\\ n_C=n_{CO_2}=\dfrac{39,6}{44}=0,9mol\\ m_C=0,9.12=10,8g\\ m_H=11,7-10,8=0,9g\\ M_A=2,689.29=77,981g/mol\)
Ta có tỉ lệ
\(\dfrac{12x}{10,8}=\dfrac{y}{0,9}=\dfrac{77,981}{11,7}\\ \Rightarrow x\approx6;y\approx6\\ \Rightarrow CTPT\left(A\right):C_6H_6\\ 2C_6H_6+15O_2\rightarrow12CO_2+6H_2O\\ n_{H_2O}=\dfrac{0,9.6}{12}=0,45mol\\ m=m_{H_2O}=0,45.18=8,1g\)

giúp mình với ạ