the tich khong khi o dieu kien tieu chuan chua 20% the tich oxi can dung de dot chay 9.2gam ruou etylit nguyen chat
the tich khong khi o dieu kien tieu chuan chua 20% the tich oxi can dung de dot chay 9.2gam ruou etylit nguyen chat
By Sadie
By Sadie
the tich khong khi o dieu kien tieu chuan chua 20% the tich oxi can dung de dot chay 9.2gam ruou etylit nguyen chat
ta có pthh :C2H5OH +3O2 ———->2CO2 +3H2O
Tỳ lệ :0,2 mol—–0,6 mol—–0,4 mol
VCO2 :0,4 X 22,4 =tự tính
nCO2=0,6mol ——–>n KK =0,6 X 100/20 =3 MOL
Vkk =3 x 22,4 =67,2 l
5 sao nhaaa
Đáp án:
\({V_{kk}} = 67,2l\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_5}OH + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_5}OH}} = \dfrac{{9,2}}{{46}} = 0,2mol\\
{n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,6mol\\
{V_{{O_2}}} = 0,6 \times 22,4 = 13,44l\\
{V_{kk}} = 5{V_{{O_2}}} = 13,44 \times 5 = 67,2l
\end{array}\)