\(\begin{array}{l}
4)\\
n_C=\frac{26,67.90\%}{12}=2(mol)\\
C+O_2\xrightarrow{t^o}CO_2\\
Theo\,PT:\,n_{CO_2(lt)}=n_C=2(mol)\\
\to V_{CO_2(tt)}=2.22,4.(100\%-15\%)=38,08(l)\\
5)\\
n_{CO_2(tt)}=\frac{15,232}{22,4}=0,68(mol)\\
\to n_{CO_2(lt)}=\frac{0,68}{85\%}=0,8(mol)\\
CaCO_3\xrightarrow{t^o}CaO+CO_2\\
Theo\,PT:\,n_{CaCO_3(lt)}=n_{CO_2(lt)}=0,8(mol)\\
\to m_{da\,voi}=\frac{0,8.100}{95\%}≈842(g)\\
6)\\
2H_2+O_2\xrightarrow{t^o}2H_2O\\
Vi\,\frac{V_{H_2}}{2}<V_{O_2}\to Hieu\,suat\,tinh\,theo\,H_2\\
Theo\,PT:\,n_{H_2O(lt)}=n_{H_2(lt)}=\frac{6,72}{22,4}=0,3(mol)\\
\to H=\frac{4,32}{0,3.18}.100\%=80\%
\end{array}\)

Đáp án $+$ Giải thích các bước giải:

 Bài 4:

$m_C=\dfrac{26,67×90}{100}≈24(g)$

$n_C=\dfrac{24}{12}=2(mol)$

$C+O_2\xrightarrow{t^o}CO_2$

Theo PT: $n_{CO_2}=n_C=2(mol)$

$n_{CO_2\ (tt)}=\dfrac{2×(100-15)}{100}=1,7(mol)$

$→V_{CO_2}=1,7×22,4=38,08(l)$

 Bài 5:

$n_{CO_2}=\dfrac{15,232}{22,4}=0,68(mol)$

$CaCO_3\xrightarrow{t^o}CaO+CO_2$

Theo PT: $n_{CaCO_3\ (pư)}=n_{CO_2}=0,68(mol)$

$n_{CaCO_3}=\dfrac{0,68×100}{85}=0,8(mol)$

$→m_{đá\ vôi}=\dfrac{0,8×100×100}{95}≈84,211(g)$

 Bài 6:

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

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

$n_{H_2O}=\dfrac{4,32}{18}=0,24(mol)$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

Xét $\dfrac{0,3}{2}<\dfrac{0,2}{1}⇒$ Hiệu suất tính theo $H_2$

Theo PT: $n_{H_2\ (pư)}=n_{H_2O}=0,24(mol)$

$→H=\dfrac{0,24}{0,3}×100\%=80\%$