PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\left(1\right)\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\left(2\right)\)
\(n_{O_2}=\dfrac{7}{22,4}=0,3125\left(mol\right)\)
Đặt số mol C2H2 là b, số mol CH4 là a, ta có hệ phương trình sau:
\(\left\{{}\begin{matrix}22,4a+22,4b=3\\2a+2,5b=0,3125\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a=\dfrac{5}{112}\left(mol\right)\\b=\dfrac{5}{56}\left(mol\right)\end{matrix}\right.\)
\(V_{CH_4}=\dfrac{5}{112}.22,4=1\left(l\right)\)
\(V_{C_2H_2}=\dfrac{5}{56}.22,4=2\left(l\right)\)
\(\%CH_4=\dfrac{1}{3}.100\%=33,33\%\)
\(\%C_2H_2=100\%-33,33\%=66,67\%\)
b) Đính chính đề: \(V_{CO_2},V_{H_2O}\) tạo ra sau phản ứng
- \(n_{CO_2\left(1\right)}=n_{CH_4}=\dfrac{5}{112}\left(mol\right)\)
\(n_{CO_2\left(2\right)}=2.n_{C_2H_2}=2.\)\(\dfrac{5}{56}=\dfrac{5}{28}\left(mol\right)\)
\(\sum V_{CO_2}=\left(\dfrac{5}{112}+\dfrac{5}{28}\right).22,4=5\left(l\right)\)
\(n_{H_2O\left(1\right)}=2.n_{CH_4}=2.\)\(\dfrac{5}{112}=\dfrac{5}{56}\left(mol\right)\)
\(n_{H_2O\left(2\right)}=n_{C_2H_2}=\dfrac{5}{56}\left(mol\right)\)
\(\sum V_{H_2O}=\left(\dfrac{5}{56}+\dfrac{5}{56}\right).22,4=4\left(l\right)\)
a.
\(CH_4+2O_2\rightarrow CO_2+2H_2O\) (1)
x\2------x---------x\2------x
\(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\) (2)
(7-x)2\5----(7-x)----(7-x).4\5-----(7-x)2\5
Gọi x là số lít oxi pt (1)--> số lit oxi phương trình (2):7-x
Ta có: \(V_{CH_4}+V_{C_2H_2}=3\Leftrightarrow\dfrac{x}{2}+\dfrac{\left(7-x\right)2}{5}=3\Leftrightarrow x=2\Rightarrow V_{CH_4=1l};V_{C_2H_2=2l}\)
%\(\%V_{CH_4}=\dfrac{1}{3}.100\%\simeq33;33\%\Rightarrow\%C_2H_2=100\%-33,34\%=66,67\%\)
b. \(V_{CO_2}=5l;V_{H_2O}=4l\)
