Đáp án:

\({\text{V = 17}}{\text{,66352 lít}}\)

Giải thích các bước giải:

 Gọi số mol \(KMn{O_4};{\text{ KCl}}{{\text{O}}_3};{\text{ }}{{\text{K}}_2}C{r_2}{O_7}\) lần lượt là x, y, z.

\(\to 158x + 122,5y + 294z = 57,45\)

Phản ứng xảy ra:

\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)

\(KCl{O_3} + 6HCl\xrightarrow{{}}KCl + 3C{l_2} + 3{H_2}O\)

\({K_2}C{r_2}{O_7} + 14HCl\xrightarrow{{}}2KCl + 2CrC{l_3} + 3C{l_2} + 7{H_2}O\)

\(\to {n_{KCl}} = {n_{KMn{O_4}}} + {n_{KCl{O_3}}} + 2{n_{{K_2}C{r_2}{O_7}}} = x + y + 2z = \frac{{29,8}}{{39 + 35,5}} = 0,4{\text{ mol}}\)

\({n_{MnC{l_2}}} = {n_{KMn{O_4}}} = x;{\text{ }}{{\text{n}}_{CrC{l_3}}} = {n_{{K_2}C{r_2}{O_7}}} = z = 2x\)

Giải được: x=0,0633; y=0,0835; z=0,1266

\(\to {n_{C{l_2}}} = \frac{5}{2}{n_{KMn{O_4}}} + 3{n_{KCl{O_3}}} + 3{n_{{K_2}C{r_2}{O_7}}} = 2,5x + 3y + 3z = 0,78855{\text{ mol}} \to {\text{V = 0}}{\text{,78855}}{\text{.22}}{\text{,4 = 17}}{\text{,66352 lit}}\)

Đáp án:

Gọi số mol $KMn{O_4}; KClO_3;K_2Cr_2O_7$ lần lượt là x, y, z.

\[\to 158x + 122,5y + 294z = 57,45\qquad(1)\]

\[2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\] \[KCl{O_3} + 6HCl\xrightarrow{{}}KCl + 3C{l_2} + 3{H_2}O\] \[{K_2}C{r_2}{O_7} + 14HCl\xrightarrow{{}}2KCl + 2CrC{l_3} + 3C{l_2} + 7{H_2}O\]

\[n_{KCl}=\dfrac{29,8}{74,5}=0,4(mol)\]

\[\to x+y+2z=0,4\qquad(2)\]

\[\left\{\begin{matrix}
n_{MnCl_2}=x(mol) & \\ 
n_{CrCl_3}=2z(mol) & 

\end{matrix}\right.\]

\[\to \dfrac{x}{2z}=\dfrac{1}{2}\to x-z=0\qquad(3)\]

Từ $(1);(2);(3)\to \left\{\begin{matrix}
x=0,1(mol) & \\ 
y=0,1(mol) & \\
z=0,1(mol) 
\end{matrix}\right.$

$\to n_{Cl_2}=2,5x+3y+3z=2,5.0,1+3.0,1+3.0,1=0,85(mol)$

$\to V=0,85\times 22,4=19,04(l)$
$\to A$