Đáp án:

4. $N_2O_5$                 

5. $NO_2$

6. a. $5,04\text{(l)}$ b. $7,2\text{(g)}$

7. 

b. $6,2\text{(g)}$

c. $\%m_C≈16,2\%$

$\%m_P≈83,8\%$

Giải thích các bước giải:

4. CTHH cần tìm có dạng \(N_xO_y\; (x;y\in\mathbb N^*)\)

\(\to 14x+16y=108\qquad(1)\)

Vì $m_O: m_N=20:7\to \dfrac{16y}{14x}=\dfrac{20}7\to \dfrac xy=\dfrac 25\to 5x-2y=0\qquad (2)$

Từ $(1);(2)\to \begin{cases}x = 2\\ y=5\end{cases}$

$\to N_2O_5$

5. CTHH cần tìm có dạng \(N_xO_y\; (x;y\in\mathbb N^*)\)

`%m_N=30,43%\to %m_O=100%-30,43%=69,57%`

`\to \frac{14x}{16y}=\frac{30,43%}{69,57%}\to x/y=1/2`

`\to NO_2`

6.

a. \(n_{KMnO_4}=\dfrac{71,1}{158}=0,45\text{(mol)}\)

PTHH: \(2KMnO_4\xrightarrow{t^{\circ}} K_2MnO_4+MnO_2+O_2(\uparrow)\)

Theo PTHH: \(n_{O_2}=\dfrac 12\cdot n_{KMnO_4}=\dfrac 12\cdot 0,45=0,225\text{(mol)}\)

\(\to V_{O_2}=0,225\cdot 22,4=5,04\text{(l)}\)

b. \(S+O_2\xrightarrow{t^{\circ}} SO_2\)

Theo PTHH: \(n_S=n_{O_2}=0,225\text{(mol)}\)

$\to m_S=0,225\cdot 32=7,2\text{(g)}$

7. 

a. PTHH: \(C+O_2\xrightarrow{t^{\circ}}CO_2\qquad(1)\\ 4P+5O_2\xrightarrow{t^{\circ}}2P_2O_5\qquad(2)\)

b. \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\text{(mol)}\)

Theo PTHH $(2)\to n_P=2\cdot n_{P_2O_5}=0,1\cdot 2=0,2\text{(mol)}$

$\to m_P=0,2\cdot 31=6,2\text{(g)}$

c. $n_{CO_2}=\dfrac{2,24}{22,4}=0,1\text{(mol)}$

Theo PTHH $(1)\to n_C=n_{CO_2}=0,1\text{(mol)}$

$\to m_C=0,1\cdot 12=1,2\text{(g)}$

$\to \%m_C=\dfrac{1,2}{1,2+6,2}\cdot 100\%≈ 16,2\%$

$\to \%m_P≈100\%-16,2\%≈83,8\%$