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PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)
\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,15\left(mol\right)\\n_{KMnO_4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{KMnO_4}=0,3\cdot158=47,4\left(g\right)\end{matrix}\right.\)
\(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,2--->0,4--------->0,2
\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=0,4.22,4=8,96\left(l\right)\\m_{CO_2}=0,2.44=8,8\left(g\right)\end{matrix}\right.\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(0.2..........0.15\)
\(V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(0.3...............................................0.15\)
\(m_{KMnO_4}=0.3\cdot158=47.4\left(g\right)\)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
b, Sửa đề: 17,9 (l) → 17,92 (l)
Ta có: \(n_{CO_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)=n_C\)
\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\Rightarrow n_H=1.2=2\left(mol\right)\)
⇒ mA = mC + mH = 0,8.12 + 2.1 = 11,6 (g)
Theo ĐLBT KL, có: mA + mO2 = mCO2 + mH2O
⇒ mO2 = 0,8.44 + 18 - 11,6 = 41,6 (g)
\(\Rightarrow n_{O_2}=\dfrac{41,6}{32}=1,3\left(mol\right)\Rightarrow V_{O_2}=1,3.22,4=29,12\left(l\right)\)
16nmetan+58nbutan=7,4 (1).
BT C: nmetan+4nbutan=22/44=0,5 (2).
Giải hệ phương trình gồm (1) và (2), ta suy ra nmetan=0,1 (mol) và nbutan=0,1 (mol).
Số mol nước tạo ra là 0,5.(0,1.4+0,1.10)=0,7 (mol).
BTKL: 7,4+32nkhí oxi=22+0,7.18, suy ra nkhí oxi=0,85 (mol).
Thể tích khí oxi cần tìm là 0,85.22,4=19,04 (lít).
\(n_{CO_2}=\dfrac{22}{44}=0,5mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_4H_{10}}=y\end{matrix}\right.\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x 2x x ( mol )
\(2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\)
y 13/2 y 4y ( mol )
Ta có:
\(\left\{{}\begin{matrix}16x+58y=7,4\\x+4y=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow n_{O_2}=2.0,1+\dfrac{13}{2}.0,1=0,85mol\)
\(V_{O_2}=0,85.22,4=19,04l\)
a) \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,05-->0,1------->0,05
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,125<--0,3125<----0,25
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,05+0,125}.100\%=28,57\%\\\%V_{C_2H_2}=\dfrac{0,125}{0,05+0,125}.100\%=71,43\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{0,05.16+0,125.26}.100\%=19,753\%\\\%m_{C_2H_2}=\dfrac{0,125.26}{0,05.16+0,125.26}.100\%=80,247\%\end{matrix}\right.\)
b) \(n_{O_2}=0,1+0,3125=0,4125\left(mol\right)\)
=> \(V_{O_2}=0,4125.22,4=9,24\left(l\right)\)
=> Vkk = 9,24.5 = 46,2 (l)
\(a)\\ 4Al+ 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ \)
b) Bảo toàn khối lượng :
\(m_{O_2} = 21,8 -13,8 =8(gam)\\ n_{O_2} = \dfrac{8}{32} = 0,25(mol)\\ V_{O_2} = 0,25.22,4 = 5,6(lít)\)
c)
\(n_{Al} = a(mol) ; n_{Fe} = b(mol)\Rightarrow 27a + 56b = 13,8(1)\\ n_{O_2} = 0,75a + \dfrac{2}{3}b = 0,25(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,15\\ \%m_{Al} = \dfrac{0,2.27}{13,8}.100\% =39,13\%\\ \%m_{Fe} = 100\% -39,13\% = 60,87\%\)
\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(0.05.0.05...0.05\)
\(\Rightarrow Sdư\)
\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(b.\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(0.1..0.1\)
\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
b, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{6.10^{22}}{6.10^{23}}=0,1\left(mol\right)\)
Theo PT: \(n_{O_2}=2n_{CH_4}=0,2\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)
c, Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PT: \(S+O_2\underrightarrow{t^o}SO_2\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PT: \(n_{O_2}=n_S+\dfrac{5}{4}n_P=0,35\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,35.22,4=7,84\left(l\right)\)