Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$
b)
n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)
m dd NaHCO3 = 0,2.84/8% = 210(gam)
c)
n CO2 = n CH3COOH = 0,2(mol)
=> V CO2 = 0,2.22,4 = 4,48(lít)
d)
m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100 + 210 - 0,2.44 = 301,2(gam)
C% CH3COONa = 0,2.82/301,2 .100% = 5,44%
PT \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+H_2O+CO_2\uparrow\)
a, \(m_{CH_3COOH}=\frac{6\times150}{100}=9\left(g\right)\)\(\Rightarrow n_{CH_3COOH}=\frac{9}{60}=0,15\left(mol\right)\)
Theo PT \(n_{NaHCO_3}=n_{CH_3COOH}=0,15\left(mol\right)\Rightarrow m_{NaHCO_3}=0,15\times84=12,6\left(g\right)\)\(\Rightarrow m_{ddNaHCO_3}=\frac{12,6\times100}{8,4}=150\left(g\right)\)
b, Theo PT \(n_{CH_3COONa}=n_{CH_3COOH}=0,15\left(mol\right)\)
\(\Rightarrow m_{CH_3COONa}=0,15\times82=12,3\left(g\right)\)
Có \(m_{ddsaupu}=150+150=300\left(g\right)\)
\(\Rightarrow C\%=\frac{12,3}{300}\times100=4,1\%\)
Khối lượng dung dịch NaHCO3:
CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O
nCH3COOH = 12 / 60 = 0,2 mol.
nNaHCO3 = 0,2mol.
mNaHCO3 cần dùng = 0,2 x 84 = 16,8g.
Khối lượng dung dịch NaHCO3 cần dùng: 16,8 x 100 / 8,4 = 200g.
Gọi nNa2CO3 = x (mol)
Na2CO3 + 2HCl \(\rightarrow\) 2NaCl + H2O + CO2
x \(\rightarrow\) 2x \(\rightarrow\) 2 x (mol)
C%(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%
=> x =0,547 (mol)
mNa2CO3 = 0,547 . 106 = 57,982 (g)
mHCl = 2 . 0,547 . 36,5 =39,931 (g)
C%(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%
C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%
Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)
a, \(C_{M_{CH_3COOH}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)
CaCO3 + 2 CH3COOH -> (CH3COO)Ca + CO2 + H2O
b) nCaCO3=0,1(mol)
=> nCH3COOH = 0,2(mol)
=> mCH3COOH= 0,2. 60=12(g)
=> mddCH3COOH=(12.100)/12=100(g)
CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O
mCH3COOH = 150 x 6/100 = 9 (g)
==> nCH3COOH = m/M = 9/60 = 0.15 (mol)
Theo phương trình => nNaHCO3 = 0.15 (mol)
mNaHCO3 = n.M = 84 x 0.15 = 12.6 (g)
==> mddNaHCO3 = 12.6x100/8.4 = 150 (g)
mdd sau pứ = 150 + 150 = 300 (g)
mCH3COONa = n.M = 0.15 x 82 = 12.3 (g)
C%dd muối sau pứ = 12.3 x 100/300 = 4.1 (%)
\(m_{CH_3COOH}=\dfrac{80.9}{100}=7,2\left(g\right)\)
\(n_{CH_3COOH}=\dfrac{7,2}{60}=0,12\left(mol\right)\)
PTHH :
\(15CH_3COOH+10NaHCO_3\rightarrow10CH_3COONa+2H_2O+20CO_2\uparrow\)
0,12 0,08 0,08 0,016 0,16
\(a,m_{NaHCO_3}=84.0,08=6,72\left(g\right)\)
\(m_{ddNaHCO_3}=\dfrac{6,72.100}{4,2}=160\left(g\right)\)
\(b,m_{CH_3COONa}=0,08.82=6,56\left(g\right)\)
\(m_{H_2O}=0,016.18=0,288\left(g\right)\)
\(m_{CO_2}=0,16.44=7,04\left(g\right)\)
\(m_{ddCH_3COONa}=80+160-0,288-7,04=232,672\left(g\right)\)
\(C\%=\dfrac{6,56}{232,672}\approx2,82\%\)
tại sao lại trừ KL nước?