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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
=> \(\left\{{}\begin{matrix}n_{O_2}=1,6-0,2-1=0,4\left(mol\right)\\n_{CO_2}=0,8\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{0,4}{0,4+0,8}.100\%=33,33\%\\\%V_{CO_2}=\dfrac{0,8}{0,4+0,8}.100\%=66,67\%\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{O_2}=\dfrac{0,4.32}{0,4.32+0,8.44}.100\%=26,67\%\\\%m_{CO_2}=\dfrac{0,8.44}{0,4.32+0,8.44}.100\%=73,33\%\end{matrix}\right.\)
Gọi $n_{Mg} = a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b = 10,35(1)$
$2Mg + O_2 \xrightarrow{t^o} 2MgO$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$n_{O_2} = \dfrac{1}{2}a + \dfrac{3}{4}b = \dfrac{5,88}{22,4} = 0,2625(2)$
Từ (1)(2) suy ra a = 0,15 ; b = 0,25
$m_{Mg} = 0,15.24 = 3,6(gam)$
$m_{Al} = 0,25.27 = 6,75(gam)$
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
Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO}=b\left(mol\right)\end{matrix}\right.\)⇒ 2a + 28b = 6,8(1)
\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ 2CO + O_2 \xrightarrow{t^o} 2CO_2\)
Theo PTHH :
\(n_{O_2} = 0,5a + 0,5b = \dfrac{8,96}{22,4} = 0,4(2)\)
Từ (1)(2) suy ra: a = 0,6 ; b = 0,2
Vậy :
\(\%m_{H_2} = \dfrac{0,6.2}{6,8}.100\% = 17,65\%\\ \%m_{CO} = 100\% - 17,65\% = 82,35\%\)
Cho em hỏi tại sao no2=0.5a+0.5b=0.4
tại sao viết 0.5 mà ko là 1 ạ
\(\overline{M}=14\cdot M_{H_2}=14\cdot2=28\left(\dfrac{g}{mol}\right)\)
\(n_X=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(m_X=0.2\cdot28=5.6\left(g\right)\)
\(CTchung:C_2H_x\)
\(BảotoànC:\)
\(n_{CO_2}=2\cdot n_{C_2H_x}=2\cdot n_X=2\cdot0.2=0.4\left(mol\right)\)
\(m_{CO_2}=0.4\cdot44=17.6\left(g\right)\)
Chúc em học tốt !!!
\(M_{hỗn\ hợp} = 4,5.2 = 9\\ Gọi : n_{CH_4} = a(mol) ; n_{H_2} = b(mol)\\ \Rightarrow 16a + 2b =9(a + b)\ (1) n_{O_2} = \dfrac{56}{5.22,4} = 0,5(mol)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{O_2} = 2a + 0,5b = 0,5(2)\\ (1)(2) \Rightarrow a = 0,2 ; b = 0,2\\ \Rightarrow V = (0,2 + 0,2).22,4 = 8,96(lít)\)
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)
\(n_{N_2}=\dfrac{4,2}{28}=0,15\left(mol\right)\)
=> \(\overline{M}=\dfrac{4,7}{0,15+0,25}=11,75\left(g/mol\right)\)
=> \(d_{hh/He}=\dfrac{11,75}{4}=2,9375\)