Đáp án:
`a)` `%V_{Cl_2}=40%, %V_{O_2}=60%`
`b)` `m_{Cl_2}=3,55g, m_{O_2}=2,4g`
Giải thích các bước giải:
Cho `Cl_2 , O_2` lần lượt là `a, b` `mol`
Ta có:
`∑_{\text{hh}}=\frac{m_{Cl_2}+m_{O_2}}{n_{Cl_2}+n_{O_2}}`
`=>∑_{\text{hh}}=\frac{71a+32b}{a+b}=2.23,8`
`=> \frac{71a+32b}{a+b}=47,6`
`=>71a+32b=47,6a+47,6b`
`=> 23,4a-15,6b=0`
`=> a=\frac{2}{3}b`
Ta có: `n_{\text{hh}}=a+b(mol)`
`a)` $\%V_{Cl_2}=\dfrac{a.100\%}{a+b}=\dfrac{100\%.\dfrac{2}{3}b}{\dfrac{2}{3}b+b}=\dfrac{100\%a}{\dfrac{5}{3}a}=40\%$
$\%V_{O_2}=\dfrac{b.100\%}{a+b}=\dfrac{100\%b}{\dfrac{2}{3}b+b}=\dfrac{100\%b}{\dfrac{5}{3}b}=60\%$
`b)` Ta có: `\frac{5,95}{a+b}=47,6`
`=> 47,6a+47,6b=5,95`
Mặc khác: `a=\frac{2}{3}b`
`=> \frac{2}{3}47,6b+47,6b=5,95`
`=> b=0,075(mol)`
`=> m_{O_2}=32.0,075=2,4g`
`=> m_{Cl_2}=0,075.\frac{2}{3}.71=3,55g`
