Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(PTHH:Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\uparrow\\ K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\\ n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}+n_{K_2SO_3}=0,2\\126n_{Na_2SO_3}+158n_{K_2SO_3}=28,4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%_{m_{Na_2SO_3}}=\dfrac{0,1\cdot126}{28,4}\cdot100\%\approx44\%\\ \Rightarrow\%_{m_{K_2SO_3}}=100\%-44\%=56\%\)
\(n_{HCl}=0,1\cdot2+0,1\cdot2=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{14,6}{200}\cdot100\%=7,3\%CC\)
\(n_{NaOH}=1,5.0,15=0,225mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_3COOH}=x\\n_{CH_3COOC_2H_5}=y\end{matrix}\right.\) ( mol )
\(\rightarrow60x+88y=15,6\left(g\right)\) (1)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
x x x ( mol )
\(CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\)
y y y ( mol )
\(n_{NaOH}=x+y=0,225\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,075\end{matrix}\right.\) ( mol )
\(\rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,15.60}{15,6}.100\%=57,69\%\\\%m_{CH_3COOC_2H_5}=100\%-57,69\%=42,31\%\end{matrix}\right.\)
\(m_{CH_3COONa}=\left(0,15+0,075\right).82=18,45g\)
\(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow a + b = \dfrac{3,36}{22,4} = 0,15(1) \)
\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ n_{CO_2} = a + 2b = \dfrac{4,48}{22,4} = 0,2(2)\)
Từ (1)(2) suy ra: a = 0,1 ; b = 0,05
Suy ra:
\(\%V_{CH_4} = \dfrac{0,1}{0,15}.100\% = 66,67\%\\ \%V_{C_2H_4} = 100\% - 66,67\% = 33,33\%\)
b)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,05(mol)\\ \Rightarrow m_{Br_2} = 0,05.160 = 8\ gam\)
\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a) Pt : \(2Na+2H_2O\rightarrow2NaOH+H_2|\)
2 2 2 1
a 0,5a
\(2K+2H_2O\rightarrow2KOH+H_2|\)
2 2 2 1
b 0,5b
b) Gọi a là số mol của Na
b là số mol của K
\(m_{Na}+m_K=10,1\left(g\right)\)
⇒ \(n_{Na}.M_{Na}+n_K.M_K=10,1g\)
⇒ 23a + 39b = 10,1g (1)
Theo phương trình : 0,5a + 0,5b = 0,15(2)
Từ(1),(2), ta có hệ phương trình :
23a + 39b = 10,1g
0,5a + 0,5b = 0,15
⇒ \(\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(m_{Na}=0,1.23=2,3\left(g\right)\)
\(m_K=0,2.39=7,8\left(g\right)\)
0/0Na = \(\dfrac{2,3.100}{10,1}=22,77\)0/0
0/0K = \(\dfrac{7,8.100}{10,1}=77,23\)0/0
Sửa đề: 44,4 → 4,44
a, \(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)
\(C_nH_{2n+1}COOH+Na\rightarrow C_nH_{2n+1}COONa+\dfrac{1}{2}H_2\)
b, Gọi: nCH3COOH = 2x (mol) ⇒ nCnH2n+1COOH = x (mol)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}+\dfrac{1}{2}n_{C_nH_{2n+1}COOH}=x+\dfrac{1}{2}x=0,03\)
⇒ x = 0,02 (mol)
⇒ nCH3COOH = 0,04 (mol), nCnH2n+1COOH = 0,02 (mol)
\(\Rightarrow0,04.60+0,02.\left(14n+46\right)=4,44\Rightarrow n=4\)
Vậy: B là C4H9COOH.
c, \(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,04.60}{4,44}.100\%\approx54,05\%\\\%m_{C_4H_9COOH}\approx45,95\%\end{matrix}\right.\)
2C2H5OH+Na->2C2H5ONa +H2
0,3------------------------------------0,15
2CH3COOH+Na->2CH3COONa+H2
0,1-------------------------------------->0,05
NaOH+CH3COOH->CH3COONa+H2O
0,1-------0,1 mol
n khí =4,48 \22,4=0,2 mol
n NaOH=0,5.0,2=0,1 mol
=>nH2 pt2=0,05
=>n H2 pt1=0,15
=>mC2H5OH=0,3.46=13,8g
=>m CH3COOH=0,1.60=6g
\(V_{C_2H_5OH}=\dfrac{100.40}{100}=40\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=40.0,8=32\left(g\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{32}{46}=\dfrac{16}{23}\left(mol\right)\)
PTHH: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)
\(\dfrac{16}{23}\)----------------------------------->\(\dfrac{8}{23}\)
\(\rightarrow V_{H_2}=\dfrac{8}{23}.22,4=\dfrac{896}{115}\left(l\right)\)