Hòa tan 37 gam CuO vào dung dịch acetic acid CH3COOH 2M
1) Tính nồng độ mol của dung dịch muối thu được
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$n_{NaOH} = \dfrac{50.10\%}{40} = 0,125(mol)$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
Theo PTHH :
$n_{CH_3COOH} = n_{CH_3COONa} = n_{NaOH} = 0,125(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,125.60}{8\%} = 93,75(gam)$
$m_{dd\ sau\ pư} = m_{dd\ CH_3COOH} + m_{dd\ NaOH} = 143,75(gam)$
$C\%_{CH_3COONa} = \dfrac{0,125.82}{143,75}.100\% = 7,13\%$
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,n_{CH_3COOH}=2.n_{H_2}=2.0,3=0,6\left(mol\right)\\ m_{ddCH_3COOH}=\dfrac{0,6.60.100}{20}=180\left(g\right)\\ b,n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ C\%_{dd\left(CH_3COO\right)_2Zn}=\dfrac{0,3.183}{180+0,3.65-0,3.2}.100\approx27,602\%\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{MgCl_2}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,4.24=9,6\left(g\right)\)
c, \(m_{MgCl_2}=0,4.95=38\left(g\right)\)
d, Bạn bổ sung thêm thể tích dd HCl nhé.
Bài 9 :
\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05--->0,1-------->0,05
a) \(C_{MddHCl}=\dfrac{0,1}{0,1}=1\left(M\right)\)
b) \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)
c) \(C_{MCuCl2}=\dfrac{0,05}{0,1}0,5\left(M\right)\)
Câu 10 :
\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
0,05-->0,1------->0,05
\(m_{ddHCl}=\dfrac{0,1.36,5}{10\%}100\%=36,5\left(g\right)\)
\(m_{ddspu}=3,6+36,5=40,1\left(g\right)\)
\(C\%_{FeCl2}=\dfrac{0,05.127}{40,1}.100\%=15,84\%\)
\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)
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: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)
\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)
\(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
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
Sửa đề: 9,2 gam Na
\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
0,4------------------>0,8
\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)
\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
0,4----------------->0,8
\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\\ n_{CuO}=\dfrac{37}{80}\left(mol\right)=n_{\left(CH_3COO\right)_2Cu}\\ n_{CH_3COOH}=2.\dfrac{37}{80}=\dfrac{37}{40}\left(mol\right)\\ V_{ddCH_3COOH}=\dfrac{\dfrac{37}{40}}{2}=\dfrac{37}{80}\left(l\right)\\ C_{Mdd\left(CH_3COO\right)_2Cu}=\dfrac{\dfrac{37}{80}}{\dfrac{37}{80}}=1\left(M\right)\)