5. B

$n_{O_2}=\dfrac{4,48}{22,4}=0,2(mol)$
Ta có $n_{H_2O}=n_{CO_2}$⇒Este no, đơn chức, hở.
$C_nH_{2n}O_2+\dfrac{3n-2}{2}\xrightarrow{t^o}nCO_2+nH_2O$

0,1 : 0,2 (mol)
$⇒0,1.\dfrac{3n-2}{2}=0,2⇒n=2$

$⇒B.HCOOCH_3$

7. C

$V_{este}=n_{este}=2(mol)$
$V_{H_2O}=V_{CO_2}⇒n_{H_2O}=n_{CO_2}=8(mol)$

Ta có $n_{H_2O}=n_{CO_2}$⇒Este no, đơn chức, hở.
$C_nH_{2n}O_2+\dfrac{3n-2}{2}\xrightarrow{t^o}nCO_2+nH_2O$

Ta có :$n_{CO_2}=2.n=8(mol)$

=>$n=4$

⇒C

8. A

$C_nH_{2n}O_2+\dfrac{3n-2}{2}\xrightarrow{t^o}nCO_2+nH_2O$

$n_{CO_2}=\dfrac{m+2,4}{44}(mol)$

$n_{H_2O}=\dfrac{m-2,4}{18}(mol)$

Ta có: $n_{CO_2}=n_{H_2O}$

$⇔\dfrac{m+2,4}{44}=\dfrac{m-2,4}{18}$

$⇒m≈6(g)$

$⇒m_{CO_2}=6+2,8=8,8(g)⇒n_{CO_2}=0,2(mol)=n_{H_2O}$

Ta có:$\dfrac{6}{14n+32}=\dfrac{0,2}{n}$

$⇒n=2$

$⇒A$

10. C

$C_nH_{2n}O_2+\dfrac{3n-2}{2}\xrightarrow{t^o}nCO_2+nH_2O$
$CO_2+Ba(OH)_2\xrightarrow{}BaCO_3+H_2O$
$BaCO_3+CO_2+H_2O\xrightarrow{}BaHCO_3$

$n_{BaCO_3}=\dfrac{11,82}{197}=0,06(mol)$

$n_{Ba(HCO_3)_2}=\dfrac{11,655}{259}=0,045(mol)$
$n_{CO_2}=0,06+2.0,045=0,15(mol)$

( 2 C trong $Ba(HCO_3)_2$ nên phải nhân 2)

Ta có: $\dfrac{3,06}{14n+32}=\dfrac{0,15}{n}$

$⇒n=5$

$⇒C$

Câu 5: $B$

Thuỷ phân $X$ trong $NaOH$ thu 2 chất hữu cơ, đốt este thu được $n_{CO_2}=n_{H_2O}$

$\to X$ là este no, đơn, hở

$n_{O_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$n_{CO_2}=n_{H_2O}=x(mol)$

Bảo toàn $O$:

$0,1.2+0,2.2=2x+x$

$\to x=0,2$

Số $C=\dfrac{n_{CO_2}}{n_X}=2$

Vậy CTPT este: $C_2H_4O_2$, CTCT: $HCOOCH_3$

Câu 7: $C$

$V_{CO_2}=V_{H_2O}\to$ este no, đơn, hở $C_nH_{2n}O_2$

$\to n=\dfrac{V_{CO_2}}{V_A}=4$

Vậy CTPT este: $C_4H_8O_2$

Câu 8: $A$

Đốt este no đơn hở ta có: $n_{CO_2}=n_{H_2O}$

$\to \dfrac{m+2,8}{44}=\dfrac{m-2,4}{18}$

$\to m=6$

$\to n_{CO_2}=0,2(mol)$

Đặt CT este: $C_nH_{2n}O_2$

$\to n_{\rm este}=\dfrac{0,2}{n}$

$\to M=\dfrac{6n}{0,2}=30n=14n+32$

$\to n=2$ ($C_2H_4O_2$)

CTCT: $HCOOCH_3$

Câu 10: $C$

$n_{BaCO_3}=\dfrac{11,82}{197}=0,06(mol)$

$n_{Ba(HCO_3)_2}=\dfrac{11,655}{259}=0,045(mol)$

Bảo toàn $C$: $n_{CO_2}=n_{BaCO_3}+2n_{Ba(HCO_3)_2}=0,15(mol)$

Đặt CT este: $C_nH_{2n}O_2$

$\to n_X=\dfrac{0,15}{n}(mol)$ 

$\to 14n+32=\dfrac{3,06n}{0,15}$

$\to n=5$