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$M_X = 18,5.2 = 37$
Mà $M_{CO_2} = 44> M_X = 37$
Suy ra : $M_{oxit\ nito} < 37$
Gọi CTHH của oxit là $N_xO_y$
Ta có :
$14x + 16y < 37$. Với x = y = 1 thì thỏa mãn
Vậy oxit là $NO$
Gọi $n_{CO_2} = a(mol) ; n_{NO} = b(mol)$
Ta có :
$44a + 30b = 37(a + b) \Rightarrow 7a = 7b \Rightarrow a = b$
$\%V_{CO_2} = \%V_{NO} = \dfrac{1}{2}.100\% = 50\%$
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) Mhh = 13,25.2 = 26,5 (g/mol
Áp dụng sơ đồ đường chéo:
\(\dfrac{V_{CH_4}}{V_{C_2H_6}}=\dfrac{30-26,5}{26,5-16}=\dfrac{1}{3}\\ \rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1}{1+3}.100\%=25\%\\\%V_{C_2H_6}=100\%-25\%=75\%\end{matrix}\right.\)
b) \(\%H=\dfrac{4+3.6}{16+3.30}.100\%=20,75\%\)
a) Gọi nO2 =a (mol); nO3 = b(mol)
Có: \(\dfrac{32a+48b}{a+b}=20.2=40\)
=> 32a + 48b = 40a + 40b
=> 8a = 8b => a = b
=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{a}{a+b}.100\%=\dfrac{a}{a+a}.100\%=50\%\\\%V_{O_3}=100\%-50\%=50\%\end{matrix}\right.\)
b) Gọi nN2 =a (mol); nNO = b(mol)
Có: \(\dfrac{28a+30b}{a+b}=14,75.2=29,5\)
=> 28a + 30b = 29,5a + 29,5b
=> 1,5a = 0,5b
=> 3a = b
=> \(\left\{{}\begin{matrix}\%V_{N_2}=\dfrac{a}{a+b}.100\%=\dfrac{a}{a+3a}.100\%=25\%\\\%V_{NO}=100\%-25\%=75\%\end{matrix}\right.\)
1)
Coi \(n_X = 1(mol)\)
Gọi : \(n_{CO_2} = a(mol) ; n_{N_2} = b(mol)\)
Ta có :
\(n_X = a + b = 1(mol)\\ m_X = 44a + 28b = 1.1,225.32(gam)\\ \Rightarrow a = 0,7 ; b = 0,3\)
Vậy :
\(\%V_{CO_2} = \dfrac{0,7}{1}.100\% = 70\%\\ \%V_{N_2} = 100\% - 70\% = 30\%\)
2)
\(n_X = \dfrac{1}{22,4}(mol)\\ \Rightarrow m_X = n.M = \dfrac{1}{22,4}.1,225.32 = 1,75(gam)\)
1, a, + 8.2=16 => CH4
+ 8,5 . 2 = 17 => NH3
+ 16 . 2 =32 => O2
+ 22 . 2 = 44 => CO2
b, + 0,138 . 29 \(\approx4\) => He
+ 1,172 . 29 \(\approx34\) => H2S
+ 2,448 . 29 \(\approx71\Rightarrow Cl_2\)
+ 0,965 . 29 \(\approx28\) => N