Đáp án C

\(MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ n_{Cl_2} = n_{MnO_2} =\dfrac{8,7}{87} = 0,1(mol)\\ \Rightarrow V_{Cl_2} = 0,1.22,4 = 2,24(lít)\)

C nhe bn

PTHH: 

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$

Ta có: $n_{Mn{O}_2}=\frac{8,7}{87}=0,1\left(mol\right)=n_{C{l}_2}$

$\Rightarrow V_{C{l}_2}=0,1\cdot 22,4=2.24\left(l\right)$

khó hiểu quá ạ

\(n_{Cl_2}=a\left(mol\right)\)

\(2KI+Cl_2\rightarrow2KCl+I_2\)

\(........a..............a\)

\(BTKL:\)

\(m+71a=m-36.6+254a\)

\(\Rightarrow a=0.2\)

\(V_{Cl_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(Cl_2+2NaBr\rightarrow2NaCl+Br_2\\Cl_2+2NaI\rightarrow2NaCl+I_2\\ n_{NaBr}=1.0,2=0,2\left(mol\right)\\ n_{NaI}=0,2.2=0,4\left(mol\right)\\ n_{Cl_2}=\dfrac{n_{NaBr}+n_{NaI}}{2}=\dfrac{0,2+0,4}{2}=0,3\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(lít\right)\)

B

B

2Na+Cl2-to>2Nacl

0,1----0,05

n Na=0,1 mol

=>VCl2=0,05.22,4=1,12l

=>A

Gọi số mol Fe, Cu là a, b (mol)

=> 56a + 64b = 12,4 (1)

PTHH: 2Fe + 3Cl₂ --t^o--> 2FeCl₃

              a---->1,5a

             Cu + Cl₂ --t^o--> CuCl₂

              b--->b

=> 1,5a + b = \(\dfrac{5,04}{22,4}=0,225\) (2)

(1)(2) => a = 0,05 (mol); b = 0,15 (mol)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,05------------------>0,05

=> V_H2 = 0,05.22,4 = 1,12 (l)

=> C

\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\\n_{Fe}=c\left(mol\right)\end{matrix}\right.\)⇒ 24a + 27b + 56c = 26,05(1)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al +6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\)

\(Mg + Cl_2 \xrightarrow{t^o} MgCl_2\\ 2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Cl_2} = a + 1,5b + 1,5c = \dfrac{17,36}{22,4} = 0,775(3)\)

Từ (1)(2)(3) suy ra:  a = 0,325 ; b = -0,05 ; c = 0,35

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