Sau khi đốt cháy hoàn toàn, hỗn hợp A gồm $CO_2$ và $O_2$ dư.

Ta có:

$M_A=1,25\cdot M_{O_2}=1,25\cdot32=40g/mol$

Gọi $x$ là % thể tích $CO_2$.

$40=44x+32(1-x)$

$40=44x+32-32x$

$8=12x$

$x=\dfrac{2}{3}$

$%V_{CO_2}=\dfrac{2}{3}\cdot100=66,67%$

$%V_{O_2}=100-66,67=33,33%$

Khi cho A vào $Ca(OH)_2$ dư:

$CO_2+Ca(OH)_2\rightarrow CaCO_3+H_2O$

$n_{CaCO_3}=\dfrac{6}{100}=0,06mol$

$n_{CO_2}=0,06mol$

Vì $CO_2$ chiếm $66,67%$:

$n_A=\dfrac{0,06}{2/3}=0,09mol$

$n_{O_2\ dư}=0,09-0,06=0,03mol$

Theo:

$C+O_2\rightarrow CO_2$

$n_C=n_{CO_2}=0,06mol$

$m_C=0,06\cdot12=0,72g$

Số mol $O_2$ ban đầu:

$n_{O_2}=n_{O_2\ phản\ ứng}+n_{O_2\ dư}$

$n_{O_2}=0,06+0,03=0,09mol$

$V_{O_2}=0,09\cdot22,4=2,016L$

Hỗn hợp A:

$M_A=19,2\cdot2=38g/mol$

Với $M_{O_2}=32,\ M_{O_3}=48$.

Gọi $x$ là phần mol $O_3$:

$38=32(1-x)+48x$

$38=32+16x$

$x=0,375$

$%V_{O_3}=37,5%$

$%V_{O_2}=62,5%$

Hỗn hợp B:

$M_B=3,6\cdot2=7,2g/mol$

Với $M_{H_2}=2,\ M_{CO}=28$.

Gọi $x$ là phần mol $CO$:

$7,2=2(1-x)+28x$

$7,2=2+26x$

$x=0,2$

$%V_{CO}=20%$

$%V_{H_2}=80%$

Trong $1mol$ B:

$n_{H_2}=0,8mol$

$n_{CO}=0,2mol$

Phản ứng:

$2H_2+O_2\rightarrow2H_2O$

$2CO+O_2\rightarrow2CO_2$

Lượng $O_2$ cần:

$n_{O_2}=0,8\cdot\dfrac12+0,2\cdot\dfrac12$

$n_{O_2}=0,5mol$

Trong $1mol$ A có:

$n_{O_2}=0,625mol$

$n_{O_3}=0,375mol$

Mỗi $1mol\ O_3$ cung cấp lượng oxi tương đương:

$O_3\rightarrow\dfrac32O_2$

Lượng $O_2$ tương đương trong $1mol$ A:

$n_{O_2,tđ}=0,625+0,375\cdot\dfrac32$

$n_{O_2,tđ}=1,1875mol$

Gọi số mol A cần dùng là $n$:

$1,1875n=0,5$

$n=\dfrac{0,5}{1,1875}\approx0,421mol$

a) \(M_{hh}=0,6.29=17,4\) (g/mol)

Dùng phương pháp đường chéo :

CH4 : 16 C2H6 : 30 17,4 12,6 1,4 = 9 1

=> Trong 3 lít hỗn hợp trên \(\left\{{}\begin{matrix}2,7\left(lít\right)CH_4\\0,3\left(lít\right)C_2H_6\end{matrix}\right.\)

PTHH : \(CH_4+2O_2-t^o->CO_2+2H_2O\)   (1)

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

Theo pthh (1) và (2) : \(\Sigma n_{O2}=2n_{CH4}+\dfrac{7}{2}n_{C2H6}\)

=> \(\Sigma_{V_{O2}}=2V_{CH4}+\dfrac{7}{2}V_{C2H6}=6,45\left(l\right)\)

b) HD : Áp dụng ĐLBTKL : m_hh + m_o2 = m_sp.cháy

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

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\left(1\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{56}{22,4}=2,5\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,5\left(mol\right)\\n_{C_2H_2}=1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,5.22,4}{33,6}.100\%\approx33,33\%\\\%V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=3,5\left(mol\right)\Rightarrow m_{O_2}=3,5.32=112\left(g\right)\)

2CH3COOH+CaCO3-to>(CH3COO)2Ca+H2O+CO2

0,4-----------------0,2----------------------------------------0,2

2CH3COOH+CaO->(CH3COO)2Ca+H2O

0,1----------------0,05

n CO2=0,2 mol

=>%m CaCO3=\(\dfrac{0,2.100}{22,8}100=87,72\%\)

=>%m CaO=12,28%

=>n CaO=0,05 mol

=>VCH3COOH=\(\dfrac{0,5}{2}=0,25l\)

a)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: CaCO₃ + 2CH₃COOH --> (CH₃COO)₂Ca + CO₂ + H₂O

                0,2<---------0,4<------------------------------0,2

=> \(m_{CaCO_3}=0,2.100=20\left(g\right)\)

=> \(\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{20}{22,8}=87,72\%\\\%m_{CaO}=100\%-87,72\%=12,28\%\end{matrix}\right.\)

b)

\(n_{CaO}=\dfrac{22,8-20}{56}=0,05\left(mol\right)\)

PTHH: CaO + 2CH₃COOH --> (CH₃COO)₂Ca + H₂O

            0,05---->0,1

=> \(V_{dd.CH_3COOH}=\dfrac{0,1+0,4}{2}=0,25\left(l\right)\)

c) \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{a}{60}\left(mol\right)\\n_{C_2H_5OH}=\dfrac{1,5a}{46}\left(mol\right)\\n_{CH_3COOC_2H_5}=\dfrac{1,2a}{88}\left(mol\right)\end{matrix}\right.\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO_4(đ),t^o--> CH₃COOC₂H₅ + H₂O

Xét tỉ lệ: \(\dfrac{\dfrac{a}{60}}{1}< \dfrac{\dfrac{1,5a}{46}}{1}\) => HIệu suất tính theo CH₃COOH

\(n_{CH_3COOH\left(pư\right)}=\dfrac{1,2a}{88}\left(mol\right)\)

=> \(H=\dfrac{\dfrac{1,2a}{88}}{\dfrac{a}{60}}.100\%=81,82\%\)

a.\(m_{Br_2}=8g\)

\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)

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

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

0,025     0,05                      ( mol )

\(\%V_{C_2H_2}=\dfrac{0,025}{0,2}.100=12,5\%\)

\(\%V_{CH_4}=100\%-12,5\%=87,5\%\)

b.

\(m_{C_2H_2}=0,025.26=0,65g\)

\(m_{CH_4}=\left(0,2-0,025\right).16=2,8g\)

c.

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

0,025     0,0625                                      ( mol )

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

0,175   0,35                                  ( mol )

\(V_{kk}=V_{O_2}.5=\left(0,35+0,0625\right).22,4.5=46,2l\)

pt:

CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O

0,3       → 0,3              0,3

CO₂ + CaCO₃ + H₂O → Ca(HCO₃)₂

0,1       → 0,1                          0,1

m _{dd tăng} = m_CO2 + m_H2O – m_CaCO3 = 12g

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

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

\(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,25\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,25.22,4}{6,72}.100\%\approx83,33\%\\\%V_{C_2H_2}\approx16,67\%\end{matrix}\right.\)

Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)