Đáp án:
\(\begin{array}{l} a,\ V_{CH_4}\text{(dư)}=0,56\ lít.\\ b,\ V_{CO_2}=1,68\ lít.\\ m_{H_2O}=2,7\ g.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1\ mol.\\ n_{O_2}=\dfrac{3,36}{22,4}=0,15\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,1}{1}>\dfrac{0,15}{2}\\ \Rightarrow CH_4\ \text{dư.}\\ \Rightarrow n_{CH_4}\text{(dư)}=0,1-\dfrac{0,15}{2}=0,025\ mol.\\ \Rightarrow V_{CH_4}\text{(dư)}=0,025\times 22,4=0,56\ lít.\\ b,\ Theo\ pt:\ n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,075\ mol.\\ \Rightarrow V_{CO_2}=0,075\times 22,4=1,68\ lít.\\ Theo\ pt:\ n_{H_2O}=n_{O_2}=0,15\ mol.\\ \Rightarrow m_{H_2O}=0,15\times 18=2,7\ g.\end{array}\)
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