Đáp án:

 $1/$

$A:C_3H_8$

 $CTCT$:

$CH_3-CH_2-CH_3$

$B:C_4H_8$

Các $CTCT$

$CH_2=CH-CH_2-CH_3$

$CH_3-CH=CH-CH_3$

$CH_2=C(CH_3)-CH_3$

$2/$

$A:C_2H_6$

$CTCT : CH_3-CH_3$

$B:C_4H_8$

Các $CTCT$

$CH_2=CH-CH_2-CH_3$

$CH_3-CH=CH-CH_3$

$CH_2=C(CH_3)-CH_3$

Giải thích các bước giải:

1/

Khi cho hh đó vào brom dư thì chỉ có $Anken$ phản ứng

$C_nH_{2n}+Br_2 \to C_nH_{2n}Br_2$

$nBr_2=nB=\frac{32}{160}=0,2$

$nCO_2=\frac{74,8}{44}=1,7$

Ta có $\left \{ {{Ankan:C_nH_{2n+2}(n≥1)} \atop {Anken:C_mH_{2m}(m≥2)}} \right.$ 

Gọi số mol $Ankan=xmol$

Bảo toàn nguyên tố $"C"$

$⇒nnC_nH_{2n+2}+mnC_mH_{2m}=nCO_2=1,7$

$⇔nx+m0,2=nCO_2=1,7(1)$

$⇔n=\frac{1,7-0,2m}{x}$

 Ta có $mA+mB=24,4$

$⇒(14n+2)nC_nH_{2n+2}+14mnC_mH_{2m}=24,4$

$⇔14nx+2x+14m.0,2=24,4$

$⇔14x\frac{1,7-0,2m}{x}+2x+2,8m=24,4$

$⇔23,8-2,8m+2x+2,8m=24,4$

$⇔x=0,3$

Thay $x=0,3$ vào $(1)$

$⇒0,3n+0,2m=1,7$

$+m=4⇒x=3$

$⇒C_4H_8 ; C_3H_8$

$CTCT:C_4H_8$:

$CH_2=CH-CH_2-CH_3$

$CH_3-CH=CH-CH_3$

$CH_2=C(CH_3)-CH_3$

$CTCT:C_3H_8$:

$CH_3-CH_2-CH_3$

2/

Ta có $\left \{ {{Ankan:C_nH_{2n+2}(n≥1)} \atop {Anken:C_mH_{2m}(m≥2)}} \right.$ 

Ta có $nA+nB=\frac{2,24}{22,4}$

$⇒a+b=0,1(1)$

TN1:

$BTNT"C"⇒nCO_2=an+bm$

$BTNT "H"⇒2nH_2O=(2n+2)a+2mb⇔nH_2O=na+a+mb$

Ta có $mCO_2-mH_2O=7,6$

$⇔44(an+bm)-18(na+a+bm)=7,6$

$⇔44an+44bm-18an-18a-18bm=7,6$

$⇔26an+26bm-18a=7,6$

$⇔26(an+bm)-18a=7,6(2)$

TN2: 

$BTNT"C"⇒nCO_2=bn+am$

$BTNT "H"⇒2nH_2O=(2n+2)b+2ma⇔nH_2O=nb+b+ma$

Ta có $mCO_2-mH_2O=6,2$

$⇔44(bn+am)-18(nb+b+ma)=6,2$

$⇔44bn+44am-18nb-18b-18ma=6,2$

$⇔26bn+26am-18b=6,2$

$⇔26(bn+am)-18b=6,2(3)$

$(2)+(3) ⇒26(bn+am)-18b+26(an+bm)-18a=13,8$

$⇔26bn+26am+26an+26bm-18(a+b)=13,8$

$⇔26m(a+b)+26n(a+b)-18(a+b)=13,8$

Thay $(1)$ vào :

$⇒26m.0,1+26n.0,1-18.0,1=13,8$

$⇔n+m=6$

Vì cacbon trong $B>A$ và cả 2 đều ở thể khí

$⇒A:C_2H_6$

$CTCT : CH_3-CH_3$

$⇒B:C_4H_8$

Các $CTCT$

$CH_2=CH-CH_2-CH_3$

$CH_3-CH=CH-CH_3$

$CH_2=C(CH_3)-CH_3$