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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 \(m_{ddsaupu}=150+150=300\left(g\right)\)
\(\Rightarrow C\%=\frac{12,3}{300}\times100=4,1\%\)
mCH3COOH= 150*6/100=9g
nCH3COOH= 9/60=0.15 mol
CH3COOH + NaHCO3 --> CH3COONa + CO2 + H2O
0.15_________0.15__________0.15______0.15
mNaHCO3= 0.15*84=12.6g
mdd NaHCO3= 12.6*100/8.4=150g
m dung dịch sau phản ứng=mdd CH3COOH + mdd NaHCO3 - mCO2= 150+150-0.15*44==293.4g
C%CH3COONa= 12.3/293.4*100%= 4.19%
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 = 0.15 x 84 = 12.6 (g)
===> mddNaHCO3 = 12.6 x 100/8.4 = 150 (g)
mdd sau pứ = 150 + 150 - 0.3 = 299.7 (g)
mCH3COONa = n.M = 0.15 x 82 = 12.3 (g)
C%ddCH3COONa = 4.104 %
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%
\(m_{CH_3COOH}=\dfrac{80.9}{100}=7,2\left(g\right)\)
\(n_{CH_3COOH}=\dfrac{7,2}{60}=0,12\left(mol\right)\)
PTHH :
\(15CH_3COOH+10NaHCO_3\rightarrow10CH_3COONa+2H_2O+20CO_2\uparrow\)
0,12 0,08 0,08 0,016 0,16
\(a,m_{NaHCO_3}=84.0,08=6,72\left(g\right)\)
\(m_{ddNaHCO_3}=\dfrac{6,72.100}{4,2}=160\left(g\right)\)
\(b,m_{CH_3COONa}=0,08.82=6,56\left(g\right)\)
\(m_{H_2O}=0,016.18=0,288\left(g\right)\)
\(m_{CO_2}=0,16.44=7,04\left(g\right)\)
\(m_{ddCH_3COONa}=80+160-0,288-7,04=232,672\left(g\right)\)
\(C\%=\dfrac{6,56}{232,672}\approx2,82\%\)
CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O
mCH3COOH = 100x12/100 = 12 (g)
==> nCH3COOH = m/M = 12/60 = 0.2 (mol)
Theo pt: => nNaHCO3 = 0.2 (mol)
==> mNaHCO3 = n.M = 0.2x84 =16.8 (g)
==> mdd NaHCO3 = 16.8x100/8.4 = 200 (g)
Ta có: nCH3COONa = 0.2 (mol)
==> mCH3COONa = n.M = 0.2 x 82 = 16.4 (g)
mdd sau pứ = 200 + 100 - 0.2 x 44 =291.2 (g)
C% = 16.4 x 100/ 291.2 = 5.63%
PTHH: \(CH_3COOH+KHCO_3\rightarrow CH_3COOK+H_2O+CO_2\uparrow\)
a) Ta có: \(n_{CH_3COOH}=\dfrac{200\cdot24\%}{60}=0,8\left(mol\right)=n_{KHCO_3}\)
\(\Rightarrow m_{ddKHCO_3}=\dfrac{0,8\cdot100}{16,8\%}\approx476.2\left(g\right)\)
b) Theo PTHH: \(n_{CH_3COOK}=0,8\left(mol\right)=n_{CO_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COOK}=0,8\cdot98=78,4\left(g\right)\\m_{CO_2}=0,8\cdot44=35,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddCH_3COOH}+m_{ddKHCO_3}-m_{CO_2}=641\left(g\right)\)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{78,4}{641}\cdot100\%\approx12,23\%\)
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 (%)
tác dụng vừa đủ với dd NaHCO3 8.4%