Cho 8,6765 lít hỗn hợp khí X (đkc) gồm Cl2 và O2 phản ứng vừa đủ với 11,1 gam hỗn hợp Y gồm

Đáp án đúng là: B

$n_X\text{ = 0,35 mol; }\stackrel{BTKL}{\to }{\text{ m}}_X\text{ = 30,1 }-\text{ 11,1 = 19 gam}$

$\left.\begin{array}{l}Mg\text{ (a) }\stackrel{}{\to }{\text{ Mg}}^{+2}\text{ + 2e (2a) }\\ Al\text{ (b) }\stackrel{}{\to }{\text{ Al}}^{+3}\text{ + 3e (3b)}\end{array}\right|\left|\begin{array}{l}C{l}_2\text{ (x) + 2.1e (2x) }\stackrel{}{\to }2C{l}^{-}\\ O_2\text{ (y) + 2.2e (4y) }\stackrel{}{\to }2O^{-2}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}\stackrel{n_X}{\to }\text{ x + y = 0,35}\\ \stackrel{m_X}{\to }\text{ 71x + 32y = 19}\end{array}\right.\stackrel{}{\to }\left\{\begin{array}{l}\text{x = 0,2}\\ \text{y = 0,15}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}\stackrel{m_Y}{\to }\text{ 24a + 27b = 11,1}\\ \stackrel{BT\text{ e}}{\to }\text{ 2a + 3b = 0,2.2 + 0,15.4}\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{x = 0,35}\\ \text{y = 0,1}\end{array}\right.\stackrel{}{\to }{\text{ \%m}}_{Al}\text{ = 24,32\%}$