Đề kiểm tra cuối học kì II

\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ a,n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\Rightarrow n_{CO_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{50}{200}.0,2=0,05\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,05}{1}\Rightarrow Rượu.dư\\ \Rightarrow n_{este\left(LT\right)}=n_{axit}=0,05\left(mol\right)\\ \Rightarrow n_{este\left(TT\right)}=80\%.0,05=0,04\left(mol\right)\\ m_{CH_3COOC_2H_5}=88.0,04=3,52\left(g\right)\)

Sửa đề: 36 → 3,6

\(C_{12}H_{22}O_{11}+H_2O\underrightarrow{^{t^o,H^+}}C_6H_{12}O_6+C_6H_{12}O_6\)

Gọi: \(\left\{{}\begin{matrix}n_{C_6H_{12}O_6}=x\left(mol\right)\\n_{C_{12}H_{22}O_{11}}=y\left(mol\right)\end{matrix}\right.\) ⇒ 180x + 342y = 3,51 (1)

Theo PT: \(n_{C_6H_{12}O_6}=n_{C_{12}H_{22}O_{11}}=y\left(mol\right)\)

⇒ 180x + 180y + 180y = 3,6 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,01\left(mol\right)\\y=0,005\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_6H_{12}O_6}=\dfrac{0,01.180}{3,51}.100\%\approx51,28\%\\\%m_{C_{12}H_{22}O_{11}}\approx48,72\%\end{matrix}\right.\)

\(n_C=n_{CO_2}=0,2\left(mol\right);n_H=2n_{H_2O}=2.\dfrac{7,2}{18}=0,8\left(mol\right)\\ M_A=16.M_{H_2}=16.2=32\left(\dfrac{g}{mol}\right)\\ m_C+m_H=0,2.12+0,8.1=3,2\left(g\right)< 6,42\left(g\right)\\ m_O=6,4-3,2=3,2\left(g\right)\\ n_O=\dfrac{3,2}{16}=0,2\left(mol\right)\\ Ta.có:n_C:n_H:n_O=0,2:0,8:0,2=1:4:1\\ \Rightarrow CTTQ.A:\left(CH_4O\right)_m\left(m:nguyên,dương\right)\\ \Leftrightarrow32m=32\\ \Leftrightarrow m=1\\ Vậy.CTPT.A:CH_4O\\ CTCT:CH_3-OH\)

`C_6 H_12 O_6 -\text{lên men}-> 2C_2 H_5 OH+2CO_2 \uparrow`

        `0,3`                                 `0,6`                `0,6`       

`n_[CO_2]=[13,8]/[22,4]=0,6(mol)`

`a)C_[M_[C_6 H_12 O_6]]=[0,3]/[0,3]=1(M)`

`b)m_[C_2 H_5 OH]=0,6.46=27,6(g)`

`c)m_[C_6 H_12 O_6 (bđ)]=0,3. 100/95 .180=56,84(g)`

$n_{CO_2} = 0,1(mol) ; n_{H_2O} = 0,1(mol)$

Bảo toàn nguyên tố với C, H : 

$n_C = n_{CO_2} = 0,1(mol)$
$n_H = 2n_{H_2O} = 0,2(mol)$

$\Rightarrow n_O = \dfrac{6 - 0,1.12 - 0,2.1}{16} = 0,2875(mol)$
$n_C : n_H : n_O = 0,1 : 0,2 : 0,2875 = 1 : 2 : 2,875$

(Sai đề)

\(C_6H_{12}O_6\rightarrow^{men\text{r}ượu}_{t^0}2C_2H_5OH+2CO_2\)

\(C_2H_5OH+O_2\rightarrow^{men\text{gi}ấm}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH\rightarrow^{H_2SO_4đặc}_{t^0}CH_3COOC_2H_5+H_2O\)

\(CH_3COOC_2H_5+H_2O\rightarrow^{H_2SO_4loãng}_{t^0}CH_3COOH+C_2H_5OH\)

\(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\)