Trung hòa dung dịch acetic acid CH3COOH 8% bằng 50g dung dịch Sodium hydroxide NaOH 10% (vừa đủ). Tính khối lượng dung dịch CH3COOH đã dùng và nồng độ phần trăm của dung dịch thu được sau phản ứng.
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a, \(n_{K_2CO_3}=\dfrac{2,76}{138}=0,02\left(mol\right)\)
PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)
Theo PT: \(n_{CH_3COOH}=2n_{K_2CO_3}=0,04\left(mol\right)\)
\(\Rightarrow C\%_{CH_3COOH}=\dfrac{0,04.60}{50}.100\%=4,8\%\)
b, \(C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)
a) \(n_{CH_3COOH}=0,1.0,3=0,03\left(mol\right)\)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,03---->0,03--------->0,03
=> \(V_{dd.NaOH}=\dfrac{0,03}{1,5}=0,02\left(l\right)\)
b) mCH3COONa = 0,03.82 = 2,46 (g)
c) \(C_{M\left(CH_3COONa\right)}=\dfrac{0,03}{0,1+0,02}=0,25M\)
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
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)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,n_{CH_3COOH}=2.n_{H_2}=2.0,3=0,6\left(mol\right)\\ m_{ddCH_3COOH}=\dfrac{0,6.60.100}{20}=180\left(g\right)\\ b,n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ C\%_{dd\left(CH_3COO\right)_2Zn}=\dfrac{0,3.183}{180+0,3.65-0,3.2}.100\approx27,602\%\)
Bài 8: Bạn bổ sung thêm đề phần này nhé.
Bài 9: Bài này giống bài 2 bên dưới nhé.
Bài 10:
\(n_{Fe\left(NO_3\right)_3}=0,3.1=0,3\left(mol\right)\)
PT: \(Fe\left(NO_3\right)_3+3NaOH\rightarrow3NaNO_3+Fe\left(OH\right)_3\)
a, \(n_{NaOH}=3n_{Fe\left(NO_3\right)_3}=0,9\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,9}{2}=0,45\left(l\right)\)
b, \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{Fe\left(NO_3\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,15.160=24\left(g\right)\)
Bài 11:
Ta có: \(n_{NaOH}=\dfrac{200.12\%}{40}=0,6\left(mol\right)\)
PT: \(2NaOH+FeCl_2\rightarrow2NaCl+Fe\left(OH\right)_2\)
a, \(n_{FeCl_2}=n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,3\left(mol\right)\Rightarrow C\%_{FeCl_2}=\dfrac{0,3.127}{100}.100\%=38,1\%\)
b, \(n_{NaCl}=n_{NaOH}=0,6\left(mol\right)\)
Ta có: m dd sau pư = 200 + 100 - 0,3.90 = 273 (g)
\(\Rightarrow C\%_{NaCl}=\dfrac{0,6.58,5}{273}.100\%\approx12,86\%\)
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: \(m_{dd}=m_{ddCH_3COOH}+m_{ddKHCO_3}-m_{CO_2}=641\left(g\right)\)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{78,4}{641}\cdot100\%\approx12,23\%\)
\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)
\(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
1 2 1 1
\(m_{\left(muối\right)}=1.182=182\left(g\right)\)
\(mCH_3COOH=2.60=120\left(g\right)\)
sao có 100g dd axit mà tới 120g CH3COOH ta
$n_{NaOH} = \dfrac{50.10\%}{40} = 0,125(mol)$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
Theo PTHH :
$n_{CH_3COOH} = n_{CH_3COONa} = n_{NaOH} = 0,125(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,125.60}{8\%} = 93,75(gam)$
$m_{dd\ sau\ pư} = m_{dd\ CH_3COOH} + m_{dd\ NaOH} = 143,75(gam)$
$C\%_{CH_3COONa} = \dfrac{0,125.82}{143,75}.100\% = 7,13\%$