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pt CH3COOH+NaOH\(\rightarrow\)CH3COONa+H2O
ta có mCH3COOH=15*120:100=18 g\(\Rightarrow\)nCH3COOH=0,3 mol
mNaOH=20*100:100=20g\(\Rightarrow\)nNaOH=0,5 mol
theo pt thì NaOH dư
ta có nCH3COONa=nCH3COOH=0,3 mol\(\Rightarrow\)mCH3COONa=24,6 g
ta có m dd sau phản ứng =120+100=220 g \(\Rightarrow\)C%CH3COONa=11,2%
Mg+2CH3COOH->(CH3COO)2Mg+H2
0,15------0,3-------------0,15-------------0,15
n Mg=\(\dfrac{3,6}{24}\)=0,15 mol
m CH3COOH=24g =>n CH3COOH=\(\dfrac{24}{60}\)=0,6 mol
->CH3COOH dư
=>C% (CH3COO)2Mg=\(\dfrac{0,15.142}{200+3,6-0,15.2}\).100=10,48%
=>C% CH3COOH dư= \(\dfrac{0,3.60}{200+3,6-0,15.2}\).100=8,85%
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,2--------->0,2------------>0,2
\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)
`=>` Gợi ý:
`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)
a,Giả sử mddHCl = 36,5 (g) \(\Rightarrow n_{HCl}=\dfrac{36,5.0,2}{36,5}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: a 2a a a
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: b 2b b b
Ta có: \(2a+2b=0,2\Leftrightarrow a+b=0,1\left(mol\right)\)
mdd D = 56a+24b+36,5-(a+b).2 = 56a+24b+36,3 (g)
\(C\%_{ddFeCl_2}=\dfrac{127a.100\%}{56a+24b+36,3}=15,757\%\)
\(\Leftrightarrow127a=8,82392a+3,78168b+5,719791\)
\(\Leftrightarrow118,17608a-3,78168b=5,719791\)
Ta có: \(\left\{{}\begin{matrix}a+b=0,1\\118,17608a-3,78168b=5,719791\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,05\end{matrix}\right.\)
\(C\%_{ddMgCl_2}=\dfrac{95.0,05.100\%}{56.0,05+24.0,05+36,3}=11,79\%\)
$n_{NaOH} = \dfrac{50.10\%}{40} = 0,125(mol)$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
Theo PTHH :
$n_{CH_3COOH} = n_{CH_3COONa} = n_{NaOH} = 0,125(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,125.60}{8\%} = 93,75(gam)$
$m_{dd\ sau\ pư} = m_{dd\ CH_3COOH} + m_{dd\ NaOH} = 143,75(gam)$
$C\%_{CH_3COONa} = \dfrac{0,125.82}{143,75}.100\% = 7,13\%$
a) nMg= 3,6/24=0,15(mol)
nHCl= (300.7,3%)/36,5= 0,6(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
Ta có: 0,15/1 < 0,6/2
=> HCl dư, Mg hết => Tính theo nMg.
nY=nH2=nMgCl2=nMg=0,15(mol)
=>V(Y,đktc)=V(H2,đktc)=0,15.22,4=3,36(l)
b) mMgCl2=0,15.95= 14,25(g)
nHCl(dư)= 0,6- 0,15.2=0,3(mol)
=>mHCl(dư)=0,3.36,5= 10,95(g)
mddsau= mMg + mddHCl - mH2= 3,6+ 300 - 0,15.2= 303,3(g)
=>C%ddHCl(dư)= (10,95/303,3).100= 3,610%
C%ddMgCl2= (14,25/303,3).100=4,698%
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)
=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)
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: \(m_{dd}=m_{ddCH_3COOH}+m_{ddKHCO_3}-m_{CO_2}=641\left(g\right)\)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{78,4}{641}\cdot100\%\approx12,23\%\)
pt 2 CH3COOH+Mg\(\rightarrow\)(CH3COO)2Mg+H2
nMg=7,2:24=0,3 mol
mCH3COOH=120*20:100=24g suy ra nCH3COOH=0,4 mol
theo pt thì Mg dư
theo pt n(CH3COO)2Mg=1/2nCH3COOH=0,2 mol suy ra n(CH3COO)2Mg=28,4g
mdd sau phản ứng =7,2+120-0,4=126,8 g
suy ra C%(CH3COO)Mg=22,3%