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Na2O + H2O → 2NaOH
1 1 2
0,1 0,2
a). nNa2O=\(\dfrac{6,2}{62}\)= 0,1(mol)
CM=\(\dfrac{n}{V}\)=\(\dfrac{0,1}{4}\)= 0,025M
b). Na2O + H2SO4 → Na2SO4 + H2O
1 1 1 1
0,1 0,1
mH2SO4= n.M = 0,1 . 98 = 9,8g
⇒mddH2SO4= mct=\(\dfrac{mct.100\%}{C\%}\)= \(\dfrac{9,8.100}{20}\)= 49(g).
$CaO + H_2O \to Ca(OH)_2$
$n_{Ca(OH)_2} = n_{CaO} = \dfrac{0,28}{56} =0,005(mol)$
$C_{M_{Ca(OH)_2}} = \dfrac{0,005}{0,25} = 0,02M$
CaO+H2O→Ca(OH)2CaO+H2O→Ca(OH)2
nCa(OH)2=nCaO=0,2856=0,005(mol)nCa(OH)2=nCaO=0,2856=0,005(mol)
CMCa(OH)2=0,0050,25=0,02M
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)
e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)
\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.15\)
\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)
\(m_{CuO}=12\left(g\right)\)
\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)
\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)
Anh bổ sung câu c)
\(C_{MddNa_2SO_4}=\dfrac{0,25}{0,09879+0,5}=0,4175\left(M\right)\)
\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
2 1 1 1 (mol)
\(mCH_3COOH=2.60=120\left(g\right)\)
m muối = \(m\left(CH_3COO\right)_2Cu=1.182=182\left(g\right)\)
m H2O = 1.18 = 18 (g)
mdd = mddCH3COOH + m(CH3COO)2Cu + mH2O - mCuO
= 100 + 182 + 18 - 80 = 220 (g)
\(C\%_{ddCH_3COOH}=\dfrac{120.100}{220}=54,55\%\)
a) \(n_{CuO}=\dfrac{80}{80}=1\left(mol\right)\)
PTHH: CuO + 2CH3COOH ---> (CH3COO)2Cu + H2O
1---->2--------------------->1
=> mmuối = 1.182 = 182 (g)
b) \(C\%_{CH_3COOH}=\dfrac{60.2}{100}.100\%=120\%\) đề có sai không vậy bạn ?
Giả sử có 100 g dung dịch acid.
\(n_{MO}=n_{MSO_4}=n_{H_2SO_4}=\dfrac{a}{98}\left(mol\right)\\ m_{ddsau}=\dfrac{\left(M+16\right)a}{98}+100=\dfrac{\left(M+96\right)a}{98\cdot\dfrac{b}{100}}=\dfrac{a\left(M+96\right)}{0,98b} \)
\(\dfrac{M+16}{98}+100=\dfrac{M+96}{0,98b}\\ M+16+9800=\dfrac{100M+9600}{b}\\ bM+9816=100M+9600\\ M\left(100-b\right)=216\\ M=\dfrac{216}{100-b}\left(g\cdot mol^{-1}\right)\)
PTHH : `Ba(OH)_2 + SO_2 -> BaSO_3 + H_2O`
`a)`
`600ml = 0,6l`
`n_{SO_2} = (6,72)/(22,4) = 0,3` `mol`
`n_{Ba(OH)_2} = n_{SO_2} = 0,3` `mol`
`C_{M_(Ba(OH)_2)} = (0,3)/(0,6) =0,5` `M`
`b)`
`n_{BaSO_3} = n_{SO_3} = 0,3` `mol`
`m_{BaSO_3} = 0,3 . 217 = 65,1` `gam`
`c)`
PTHH : `Ba(OH)_2 + 2HCl -> BaCl_2 + 2H_2O`
Ta có : `n_{Ba(OH)_2} = 0,3` `mol`
`n_{HCl} = 2 . n_{Ba(OH)_2} = 0,6` `mol`
`V_{HCl} = (0,6)/(3,5) = 6/35` `l`
nCaO = \(\dfrac{m}{M}=\dfrac{0.28}{40}=0,007\left(mol\right)\)
PTHH : CaO + H2O ---> Ca(OH)2
0,007 0,007
\(C_M=\dfrac{n}{V}=\dfrac{0,007}{0,25}=0,028\left(M\right)\)
\(m_{Ca\left(OH\right)_2}=n.M=0,007.74=0,518\left(g\right)\)