Đáp án:

Bài 1: $ d_{hh/kk}\approx1,2$

Bài 2:

%$m_{CO}≈$ 20,29%

%$m_{CO_2}≈$ 79,71%

Giải thích các bước giải:

Bài 1:

$n_{CO_2}=$$\dfrac{8,8}{44}=0,2(mol)$

$n_{N_2}=$$\dfrac{7}{28}=0,25(mol)$

$\rightarrow \overline{M_{hh}}=$ $\dfrac{8,8+7}{0,2+0,25}=$ $\dfrac{15,8}{0,45}\approx 35,1(g/mol)$

$\rightarrow d_{hh/kk}=$ $\dfrac{\overline{M_{hh}}}{29} \approx $ $\dfrac{35,1}{29}\approx1,2$

Bài 2:

$n_{hh}=n_{CO}+n_{CO_2}=$ $\cfrac{15,68}{22,4}=0,7(mol)(1)$

$m_{hh}=m_{CO}+m_{CO_2}=28n_{CO}+44n_{CO_2}=27,6(g)(2)$

Từ $(1); (2)$ ta có 

$\left\{\begin{matrix} n_{CO}+n_{CO_2}=0,7 & \\ 28n_{CO}+44n_{CO_2}=27,6& \end{matrix}\right.$

$\rightarrow \left\{\begin{matrix} n_{CO}=0,2(mol) & \\ n_{CO_2}=0,5(mol)& \end{matrix}\right.$

$\rightarrow m_{CO}=0,2*28=5,6(g)$

$\rightarrow$ %$m_{CO}=$ $\dfrac{5,6}{27,6}$ * 100% $≈$ 20,29%

$\rightarrow$ %$m_{CO_2}≈$ 100% - 20,29% $≈$ 79,71%