\(n_C = n_S = a\ mol\Rightarrow 12a + 32a = 22 \Rightarrow a = 0,5\)

\(C + O_2 \xrightarrow CO_2\\ S + O_2 \xrightarrow SO_2\)

Theo PTHH : 

\(n_{O_2} = n_C + n_S = 0,5 + 0,5 = 1\ mol\\ \Rightarrow m_{O_2} = 1.32 = 32(gam)\)

a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)

\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)

\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)

b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)

\(\rightarrow V_{kk}=4,48.5=11,2l\)

c. Có \(n_{Mg}=2n_{O_2}=0,2l\)

\(\rightarrow m_{Mg}=0,2.24=4,8g\)

\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)

\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)

PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\) (1)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) (2)

\(C+O_2\underrightarrow{t^o}CO_2\) (3)

Ta có: \(n_{KCl}=\dfrac{1,49}{74,5}=0,02\left(mol\right)\)

\(n_C=\dfrac{0,24}{12}=0,02\left(mol\right)\)

\(m_Y=\dfrac{1,49}{17,028\%}=8,75\left(g\right)\), Y gồm: KCl, KMnO_4 (dư), K₂MnO₄, MnO₂.

Theo PT (3): \(n_{CO_2}=n_{O_2\left(pư\right)}=n_C=0,02\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=\dfrac{0,02}{40\%}-0,02=0,03\left(mol\right)\)

⇒ Σn_O2 = 0,02 + 0,03 = 0,05 (mol)

Theo PT (1): \(n_{O_2\left(1\right)}=\dfrac{3}{2}n_{KCl}=0,03\left(mol\right)\)

\(\Rightarrow n_{O_2\left(2\right)}=0,05-n_{O_2\left(1\right)}=0,02\left(mol\right)\)

Theo PT (2): \(n_{K_2MnO_4}=n_{MnO_2}=n_{O_2\left(2\right)}=0,02\left(mol\right)\)

\(n_{KMnO_4\left(pư\right)}=2n_{O_2}=0,04\left(mol\right)\)

Mà: m_KCl + m_{KMnO4 (dư)} + m_K2MnO4 + m_MnO2 = 8,75

⇒ m_{KMnO4 (dư)} = 1,58 (g) \(\Rightarrow n_{KMnO_4\left(dư\right)}=\dfrac{1,58}{158}=0,01\left(mol\right)\)

\(\Rightarrow H\%=\dfrac{0,04}{0,04+0,01}.100\%=80\%\)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_6}=y\left(mol\right)\end{matrix}\right.\)

\(n_{H_2O}=\dfrac{14,4}{18}=0,8\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

x-------->2x----------->2x

\(C_2H_6+\dfrac{7}{2}O_2\underrightarrow{t^o}2CO_2+3H_2O\)

y-------->3,5y------------->3y

Có hệ phương trình: \(\left\{{}\begin{matrix}x+y=\dfrac{6,72}{22,4}=0,3\\2x+3y=0,8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

a

\(\%V_{CH_4}=\dfrac{0,1.22,4.100\%}{6,72}=33,33\%\)

\(\%V_{C_2H_6}=\dfrac{0,2.22,4.100\%}{6,72}=66,67\%\)

b

\(V_{O_2}=\left(2x+3,5y\right).22,4=\left(2.0,1+3,5.0,2\right).22,4=20,16\left(l\right)\)

Cho dữ liệu dư 10% như thế thì phải hỏi là V khí \(O_2\) đã lấy/ đã dùng chứ "cần lấy" là theo PTHH (không cần cho "Biết ...")

\(V_{O_2.đã.lấy}=\dfrac{20,16.\left(100+10\right)\%}{100\%}=22,176\left(l\right)\)

Làm giúp em câu b với ạ, em cảm ơn

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

              a___a______a    (mol)

            \(S+O_2\xrightarrow[]{t^o}SO_2\)

             b___b_______b   (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}12a+32b=10\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_C=0,3\cdot12=3,6\left(g\right)\\m_S=6,4\left(g\right)\\V_{khí}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)

Chọn A.

Đáp án B