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PTTQ: C3H5(OH)3 + nRCOOH ⇌ (RCOO)nC3H5(OH)3-n + nH2O
Các công thức đúng là (a), (b), (c).
→ Đáp án C
Este của glixerol với axit cacboxylic (RCOOH) có dạng (RCOO)nC3H5(OH)3-n
→ Vậy các este đó có thể là RCOOC3H5(OH)2; (RCOO)2C3H5(OH); (RCOO)3C3H5
→ Các công thức viết đúng là (1), (2), (3)
→ 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
\(\rightarrow m_T=10.282+10.890=11720\left(g\right)\\ \rightarrow a=\dfrac{11720}{1000}=11,72\left(kg\right)\\ \rightarrow b=10.92=920\left(g\right)\)
\(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\\ PTHH:\left(RCOO\right)_3C_3H_5+3NaOH\rightarrow3RCOONa+C_3H_5\left(OH\right)_3\\ \Rightarrow n_{C_3H_5\left(OH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,1\left(mol\right)\\ \Rightarrow m_{C_3H_5\left(OH\right)_3}=0,1\cdot92=9,2\left(g\right)\\ \Rightarrow m_{RCOONa}=m_{\left(RCOO\right)_3C_3H_5}+m_{NaOH}-m_{C_3H_5\left(OH\right)_3}=89+12-9,2=91,8\left(g\right)\)
\(a) 2Na + 2C_2H_5OH \to 2C_2H_5ONa + H_2\\ b) CH_3COOH + NaOH \to CH_3COONa + H_2O\\ c) (RCOO)_3C_3H_5 +3H_2O \to 3RCOOH + C_3H_5(OH)_3\\ d) C_2H_5OH + CH_3COOH \to CH_3COOC_2H_5 + H_2O\\ e) C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ d) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ g) (RCOO)_3C_3H_5 + 3NaOH \to 3RCOONa + C_3H_5(OH)_3\)
a, \(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
b, \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
c, \(C_6H_{12}O_6+2AgNO_3+3NH_3\underrightarrow{t^o}C_5H_{11}O_5COONH_4+2Ag+2NH_4NO_3\)
d, \(\left(RCOO\right)_3C_3H_5+3NaOH\underrightarrow{t^o}3RCOONa+C_3H_5\left(OH\right)_3\)
e, \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
f, \(H_2SO_4+2CH_3COONa\rightarrow2CH_3COOH+Na_2SO_4\)
Chọn B