Đáp án:

$a,\\ \%V_{CH_4}=\%V_{C_2H_2}=50\%\\ b,\\ V_{O_2}=10,08(l)\\ m_{CO_2}=13,2(g)\\ m_{H_2O}=5,4(g)$

Giải thích các bước giải:

$a,\\ n_{Br_2}=\dfrac{32}{160}=0,2(mol)\\ C_2H_2+2Br_2\xrightarrow{t^0} C_2H_2Br_4\\ \xrightarrow{\text{Từ PTHH}}\ n_{C_2H_2}=\dfrac{1}{2}.n_{Br_2}=0,1(mol)\\ V_{C_2H_2}=0,1.22,4=2,24(l)\\ V_{CH_4}=4,48-2,24=2,24(l)\\ \%V_{CH_4}=\%V_{C_2H_2}=\dfrac{2,24}{4,48}.100\%=50\%\\ b,\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1(mol)\\ CH_4+2O_2\xrightarrow{t^0} CO_2+2H_2O\\ C_2H_2+\dfrac{5}{2}O_2\xrightarrow{t^0} 2CO_2+H_2O\\ \xrightarrow{\text{Từ PTHH}}\ n_{O_2}=2.n_{CH_4}+\dfrac{5}{2}.n_{C_2H_2}=0,45(mol)\\ V_{O_2}=0,45.22,4=10,08(l)\\ \xrightarrow{\text{Từ PTHH}}\ n_{CO_2}=n_{CH_4}+2.n_{C_2H_2}=0,3(mol)\\ m_{CO_2}=0,3.44=13,2(g)\\ \xrightarrow{\text{Từ PTHH}}\ n_{H_2O}=2.n_{CH_4}+n_{C_2H_2}=0,3(mol)\\ m_{H_2O}=0,3.18=5,4(g)$

a. $C_2H_2$ + $2Br_2$ $\rightarrow$ $C_2H_2Br_4$(1)

$n_{Br_2}$= $\frac{32}{160}$= 0,2(mol)

Từ pt(1) $\Rightarrow$ $n_{C_2H_2}$= 0,1( mol)

$V_{C_2H_2}$ = 0,1x 22,4= 2,24( l)

$V_{CH_4}$= 4,48- 2,24= 2,24( l)

%$V_{C_2H_2}$ = %$V_{CH_4}$= $\frac{2,24}{4,48}$x 100= 50%

b. $CH_4$ + $2O_2$ $\xrightarrow{to}$ $CO_2$ + $2H_2O$(2)

    $2C_2H_2$ + $5O_2$ $\xrightarrow{to}$ $2CO_2$ + $2H_2O$(3)

$n_{CH_4}$= $\frac{2,24}{22,4}$= 0,1( mol) T

ừ pt(2) $\Rightarrow$ $n_{CO_2}$= 0,1( mol); $n_{H_2O}$= 0,2(mol); $n_{O_2}$= 0,2( mol)

Từ pt(3) $\Rightarrow$ $n_{CO_2}$= 0,1( mol); $n_{H_2O}$= 0,1(mol); $n_{O_2}$= 0,25( mol)

$V_{O_2}$= (0,2+ 0,25)x 22,4= 10,08( l)

$m_{CO_2}$= (0,1+ 0,1)x 44= 8,8(g)

$m_{H_2O}$= (0,2+ 0,1)x 18= 5,4(g)

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