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: $\dfrac{3,06}{14n+32}=\dfrac{0,15}{n}$
$⇒n=5$
$⇒C$
