Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PT: \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{KOH}=0,15\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)
b, \(n_{Na_2CO_3}=0,2.0,5=0,1\left(mol\right)\)
PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\), ta được Na2CO3 dư.
Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)
CH3COOH + NaOH => CH3COONa + H2O
500 ml dd = 0.5 l dd
mNaOH = 20x30/100 = 6 (g)
nNaOH = m/M = 6/40 = 0.15 (mol)
Theo phương trình => nCH3COOH = 0.15 (mol)
CM dd CH3COOH = n/V = 0.15/0.5 = 0.3M
2CH3COOH + Na2CO3 => 2CH3COONa + CO2 + H2O
nNa2CO3 = 0.5x0.2 = 0.1 (mol)
Theo phương trình ==> nCO2 = 0.1 (mol)
VCO2 = n x 22.4 = 0.1 x 22.4 = 2.24 (l)
Câu 8 :
\(n_{MgCO3}=\dfrac{42}{84}=0,5\left(mol\right)\)
Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)
1 0,5 0,5
a) \(V_{CO2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
b) \(V_{CH3COOH}=\dfrac{1}{2}=0,5\left(l\right)\)
c) Pt : \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
1 1
300ml = 0,3l
\(C_{MCH3COONa}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)
Chúc bạn học tốt
nNaOH = 0,1.2 = 0,2 (mol)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,2<--------0,2
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)
nNaOH = 0,1 . 2 = 0,2 (mol)
PTHH: CH3COOH + NaOH -> CH3COONa + H2O
nCH3COOH = nNaOH = 0,2 (mol)
CM(CH3COOH) = 0,2/0,15 = 1,33M
$a\big)$
$n_{CH_3COOH}=\dfrac{100}{1000}.1=0,1(mol)$
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
Theo PT: $n_{NaOH}=n_{CH_3COOH}=0,1(mol)$
$\to C\%_{NaOH}=\dfrac{0,1.40}{50}.100\%=80\%$
$b\big)$
$n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1(mol)$
$2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O$
Theo PT: $\begin{cases} n_{CO_2}=n_{Na_2CO_3}=0,1(mol)\\ n_{CH_3COONa}=2n_{Na_2CO_3}=0,2(mol) \end{cases}$
$\to C\%_{CH_3COONa}=\dfrac{0,2.82}{60+10,6-0,1.44}.100\%\approx 24,77\%$
a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,2--------->0,4--------------->0,4------->0,2
=> VCO2 = 0,2.22,4 = 4,48 (l)
b) \(m_{dd.CH_3COOH}=\dfrac{0,4.60}{12\%}=200\left(g\right)\)
c) mdd sau pư = 21,2 + 200 - 0,2.44 = 212,4 (g)
=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)
Pt: \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(n_{\left(CH_3COO\right)_2Zn}=\dfrac{14,2}{183}\approx0.077mol\)
Theo pt: nH2 = n(CH3COO)2Zn = 0,077mol
=> VH2 = 1,7248l
b) Theo pt: nCH3COOH = 2n(CH3COO)2Zn = 0,154 mol
=> CMCH3COOH = 0,154 : 0,25 = 0,616M
a) nNaOH = \(\dfrac{40.20}{100.40}\)= 0,2 mol ,
pt: CH3COOH + NaOH \(\rightarrow\) CH3COONa +H2O
Theo pt nCH\(_3\)COOH = n NaOH = 0,2 mol
500 ml = 0,5 l
CM (CH\(_3\)COOH) = \(\dfrac{0,2}{0,5}\)= 0,4 M
b) nNa\(_2\)CO\(_3\)= 2.0,15 = 0,3 mol
pt: Na\(_2\)CO\(_3\) + 2CH\(_3\)COOH \(\rightarrow\) 2CH\(_3\)COONa + H2O + CO2
Tỉ lệ : nNa\(_2\)CO\(_3\) : n CH\(_3\)COOH = \(\dfrac{0,3}{1}\):\(\dfrac{0,2}{2}\)= 0,3 : 0,1
0,3>0,1 ,nên trong p ứ Na\(_2\)CO\(_3\) dư
CO2 sẽ tính theo CH\(_3\)COOH
nCO\(_2\)= 0,1 mol \(\rightarrow\) VCO\(_2\) = 22,4.0,1 = 2,24 l
a) nNaOH = 40.20100.4040.20100.40= 0,2 mol ,
pt: CH3COOH + NaOH →→ CH3COONa +H2O
Theo pt nCH33COOH = n NaOH = 0,2 mol
500 ml = 0,5 l
CM (CH33COOH) = 0,20,50,20,5= 0,4 M
b) nNa22CO33= 2.0,15 = 0,3 mol
pt: Na22CO33 + 2CH33COOH →→ 2CH33COONa + H2O + CO2
Tỉ lệ : nNa22CO33 : n CH33COOH = 0,310,31:0,220,22= 0,3 : 0,1
0,3>0,1 ,nên trong p ứ Na22CO33 dư
CO2 sẽ tính theo CH33COOH
nCO22= 0,1 mol →→ VCO22 = 22,4.0,1 = 2,24 l