A

B

A

PTHH: \(\left(RCOO\right)_3C_3H_5+3NaOH\rightarrow3RCOONa+C_3H_5\left(OH\right)_3\)

Áp dụng ĐLBTKL:

\(m_{hhaxit.béo}=8,58+1,2-0,368=9,412\left(kg\right)\\ \rightarrow m_{xà.phòng}=9,412.\left(100\%-60\%\right)=3,7648\left(kg\right)\)

Đáp án A

PTHH tổng quát:

R C O O 3 C 3 H 5     +     3 N a O H     → t 0   3 R C O O N a     +     C 3 H 5 O H 3

a) Áp dụng ĐLBTKL : 

\(m_{Cb}+m_{NaOH}=m_{hhmuối}+m_{glixerol}\)

\(\Rightarrow m_{hhmuối}=8.58+1.2-0.92=8.86\left(kg\right)\)

b) \(m_{xp}=\dfrac{m_{muối}}{60\%}=\dfrac{8.86}{60\%}=14.767\left(kg\right)\)

Chọn đáp án D

a)

Quy đổi: \(\left\{{}\begin{matrix}1,6kg=1600g\\43,22kg=43220g\end{matrix}\right.\)

\(n_{NaOH}=\dfrac{1600}{40}=40\left(mol\right)\\ \xrightarrow[]{\text{BTNT Na}}n_{Na_2CO_3}=\dfrac{40}{2}=20\left(mol\right)\)

Giả sử nếu chỉ có muối \(C_nH_{2n+1}COONa\)

\(\xrightarrow[]{\text{BTNT Na}}n_{C_nH_{2n+1}COONa}=40\left(mol\right)\)

PTHH: 

\(2C_nH_{2n+1}COONa+\left(3n+1\right)O_2\xrightarrow[]{t^o}Na_2CO_3+\left(2n+1\right)CO_2+\left(2n+1\right)H_2O\)

Theo PT: \(n_{CO_2}=n_{H_2O}=\dfrac{43220}{44+18}=697,1\left(mol\right)\)

\(\xrightarrow[]{\text{BTNT C}}n=\dfrac{697,1+20}{40}=17,9275\left(mol\right)\left(1\right)\)

Giả sử nếu chỉ có muối \(C_nH_{2n-1}COONa\)

\(\xrightarrow[]{\text{BTNT Na}}n_{C_nH_{2n-1}COONa}=40\left(mol\right)\)

PTHH: 

\(2C_nH_{2n-1}COONa+\left(4n-1\right)O_2\xrightarrow[]{t^o}Na_2CO_3+\left(2n+1\right)CO_2+\left(2n-1\right)H_2O\)

Theo PT: \(n_{CO_2}-n_{H_2O}=\left(2n+1\right)-\left(2n-1\right)=2=n_{C_nH_{2n-1}COONa}\)

\(\rightarrow n_{CO_2}-n_{H_2O}=40\left(mol\right)\)

Mà \(m_{CO_2}+m_{H_2O}=44n_{CO_2}+18n_{H_2O}=43220\left(g\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}n_{CO_2}=708,7\left(mol\right)\\n_{H_2O}=668,7\left(mol\right)\end{matrix}\right.\)

\(\xrightarrow[]{\text{BTNT C}}n=\dfrac{708,7+20}{40}=18,2175\left(mol\right)\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\rightarrow17,9275< n< 18,2175\)

Mà \(n\in N\text{*}\)

\(\rightarrow n=18\)

Vậy trong CTPT của mỗi muối có 18 nguyên tử C

Vậy CTPT của 2 muối đó lần lượt là \(C_{17}H_{35}COONa,C_{17}H_{33}COONa\)

b)

Gọi \(\left\{{}\begin{matrix}n_{C_{17}H_{35}COONa}=a\left(mol\right)\\n_{C_{17}H_{33}COONa}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)

\(\xrightarrow[]{\text{BTNT Na}}n_{NaOH}=a+b=40\left(1\right)\)

PTHH:

\(2C_{17}H_{35}COONa+68O_2\xrightarrow[]{t^o}Na_2CO_3+35CO_2+35H_2O\)

a------------------------------------------------>17,5a--->17,5a

\(2C_{17}H_{33}COONa+O_2\xrightarrow[]{t^o}Na_2CO_3+35CO_2+33H_2O\)

b------------------------------------------------>17,5b--->16,5b

\(\rightarrow m_{CO_2}+m_{H_2O}=44\left(17,5a+17,5b\right)+18\left(17,5a+16,5b\right)=1085a+1067b=43220\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\rightarrow\left\{{}\begin{matrix}a=30\left(mol\right)\\b=10\left(mol\right)\end{matrix}\right.\left(TM\right)\)

+) TH₁: Axit tự do là \(C_{17}H_{35}COOH\) và muối là \(\left(C_{17}H_{33}COO\right)_3C_3H_5\)

PTHH:

\(C_{17}H_{35}COOH+NaOH\rightarrow C_{17}H_{35}COONa+H_2O\left(1\right)\)

30<----------------------------------30

\(\left(C_{17}H_{33}COO\right)_3C_3H_5+3NaOH\rightarrow3C_{17}H_{35}COONa+C_3H_5\left(OH\right)_3\left(2\right)\)

\(\dfrac{10}{3}\)<-----------------------------------------10-------------------->\(\dfrac{10}{3}\)

\(\rightarrow m_T=30.284+\dfrac{10}{3}.884=\dfrac{34400}{3}\left(g\right)\\ \rightarrow a=\dfrac{\dfrac{34400}{3}}{1000}=\dfrac{172}{15}\left(kg\right)\)

\(\rightarrow b=\dfrac{10}{3}.92=\dfrac{920}{3}\left(g\right)\)

+) TH₂: Axit tự do là \(C_{17}H_{33}COOH\) và muối là \(\left(C_{17}H_{35}COO\right)_3C_3H_5\)

PTHH:

\(C_{17}H_{33}COOH+NaOH\rightarrow C_{17}H_{33}COONa+H_2O\)

10<----------------------------------10

\(\left(C_{17}H_{35}COO\right)_3C_3H_5+3NaOH\rightarrow3C_{17}H_{35}COONa+C_3H_5\left(OH\right)_3\)

10<------------------------------------------10--------------------->10

\(\rightarrow m_T=10.282+10.890=11720\left(g\right)\\ \rightarrow a=\dfrac{11720}{1000}=11,72\left(kg\right)\\ \rightarrow b=10.92=920\left(g\right)\)

Chọn đáp án C

Đáp án D

4