cau 11 cde 28/2

Question

Co mot hh $A$ gom $C_2H_6, C_2H_2, C_3H_6$. Dot chay hoan toan $24,8$ gam hh $A$ thu duoc $28,8$ gam nuoc. Mat khac, $0,5$ mol hh $A$ tac dung vua du voi 500 gam dd $Br_2 20\%$. Thanh phan phan tram theo the tich cua mot chat trong $A$ la:

$a. 10\% b. 25\% c. 20\% d. 35\%$

viet130480 · Answer 1 · 3/3/2014 12:07:03 AM

Bình chọn tăng 1 Bình chọn giảm

$*)\begin{cases}C_2H_6:x mol \\ C_2H_2:y mol\\C_3H_6:z mol\\30x+26y+42z=24,8 gam\\18(3x+y+3z)=28,8 gam \end{cases}$

$*)\begin{cases}C_2H_6:tx mol \\ C_2H_2:ty mol\\C_3H_6:tz mol\\tx+ty+tz=0,5 mol\\n_{Br_2}=\frac{500.20}{160.100}=0,625 mol\\2ty+tz=n_{Br_2} \end{cases}$

$\Rightarrow \begin{cases}30x+26y+42z=24,8 gam \\ 3x+y+3z=1,6 mol\\t(x+y+z)=0,5 mol\\t(2y+z)=0,625 mol \end{cases}\Rightarrow \begin{cases}15x+13y+21z=12,4 \\ 3x+y+3z=1,6\\2y+z=1,25(x+y+z) \end{cases}$

$\Rightarrow \begin{cases}x=0,2 mol \\ y=0,4 mol\\z=0,2 mol \end{cases}\Rightarrow \begin{cases}\%C_2H_6=25\% \\ \%C_2H_2=50\%\\\%C_3H_6=25\%\end{cases}\Rightarrow Chon:B$

Trả lời 03-03-14 12:07 AM