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\(\overline{M}=14\cdot M_{H_2}=14\cdot2=28\left(\dfrac{g}{mol}\right)\)
\(n_X=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(m_X=0.2\cdot28=5.6\left(g\right)\)
\(CTchung:C_2H_x\)
\(BảotoànC:\)
\(n_{CO_2}=2\cdot n_{C_2H_x}=2\cdot n_X=2\cdot0.2=0.4\left(mol\right)\)
\(m_{CO_2}=0.4\cdot44=17.6\left(g\right)\)
Chúc em học tốt !!!
Gọi số mol H2, C2H2 là a, b (mol)
=> \(\left\{{}\begin{matrix}a+b=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\\overline{M}=\dfrac{2a+26b}{a+b}=0,5.28=14\left(g/mol\right)\end{matrix}\right.\)
=> a = 0,4 (mol); b = 0,4 (mol)
\(n_{O_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)
PTHH: 2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,4--->1----------->0,8
2H2 + O2 --to--> 2H2O
0,4-->0,2
=> Y gồm \(\left\{{}\begin{matrix}CO_2:0,8\left(mol\right)\\O_{2\left(dư\right)}:0,4\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{CO_2}=\dfrac{0,8}{0,8+0,4}.100\%=66,67\%\\\%V_{O_2}=\dfrac{0,4}{0,8+0,4}.100\%=33,33\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CO_2}=\dfrac{0,8.44}{0,8.44+0,4.32}.100\%=73,33\%\\\%m_{O_2\left(dư\right)}=\dfrac{0,4.32}{0,8.44+0,4.32}.100\%=26,67\%\end{matrix}\right.\)
1)
2H2 + O2 --to--> 2H2O
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
2) Gọi số mol H2, C2H2 là a, b
=> \(\left\{{}\begin{matrix}a+b=\dfrac{17,92}{22,4}=0,8\\\dfrac{2a+26b}{a+b}=0,5.28=14\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,4\\b=0,4\end{matrix}\right.\)
\(n_{O_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,4--->0,2
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,4---->1-------------->0,8
=> \(\left\{{}\begin{matrix}n_{O_2}=1,6-0,2-1=0,4\left(mol\right)\\n_{CO_2}=0,8\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{0,4}{0,4+0,8}.100\%=33,33\%\\\%V_{CO_2}=\dfrac{0,8}{0,4+0,8}.100\%=66,67\%\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{O_2}=\dfrac{0,4.32}{0,4.32+0,8.44}.100\%=26,67\%\\\%m_{CO_2}=\dfrac{0,8.44}{0,4.32+0,8.44}.100\%=73,33\%\end{matrix}\right.\)
a) \(n_{O_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
____0,25<-0,125
=> mH2 = 0,25.2 = 0,5 (g)
=> mN2 = 4,7 - 0,5 = 4,2 (g)
b)
\(n_{N_2}=\dfrac{4,2}{28}=0,15\left(mol\right)\)
=> \(\overline{M}=\dfrac{4,7}{0,15+0,25}=11,75\left(g/mol\right)\)
=> \(d_{hh/He}=\dfrac{11,75}{4}=2,9375\)
\(n_X=\dfrac{0,896}{22,4}=0,08\left(mol\right)\)
\(M_X=21.2=42\left(g\text{/}mol\right)\\ \rightarrow m_X=0,08.42=3,36\left(g\right)\)
PTHH:
\(C_3H_4+4O_2\xrightarrow[]{t^o}3CO_2+H_2O\\ 2C_3H_6+9O_2\xrightarrow[]{t^o}6CO_2+6H_2O\\ C_3H_8+5O_2\xrightarrow[]{t^o}3CO_2+4H_2O\)
Theo PTHH: \(n_C=n_{CO_2}=3n_X=3.0,08=0,24\left(mol\right)\)
\(\rightarrow V_{CO_2}=0,24.22,4=5,376\left(l\right)\)
BTNT:
\(m_H=m_X=m_C=3,36-0,24.12=0,48\left(g\right)\\ \rightarrow n_H=\dfrac{0,48}{1}=0,48\left(mol\right)\)
Theo PTHH: \(n_{H_2O}=\dfrac{1}{2}n_H=\dfrac{1}{2}.0,48=0,24\left(mol\right)\)
\(\rightarrow m_{H_2O}=0,24.18=3,42\left(g\right)\)
MZ = 22,4 . 11,964 = 268 (g/mol)
dZ/H2 = 268/2 = 134