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a, \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,05 0,1
⇒ ddA (NaOH) làm quỳ tím đổi màu xanh
b, \(C_{M_{ddNaOH}}=\dfrac{0,1}{0,5}=0,2M\)
c, \(m_{ddA}=3,1+1.500=503,1\left(g\right)\)
d, \(C\%_{ddNaOH}=\dfrac{0,2.40.100\%}{503,1}=1,59\%\)
a, \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
b, \(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaO}=0,2\left(mol\right)\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
c, \(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,2.98}{15\%}=\dfrac{392}{3}\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{\dfrac{392}{3}}{1,05}\approx124,44\left(ml\right)\)
a, Ta có: 27nAl + 56nFe = 22 (1)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)
b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)
\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)
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: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
Na2O+H2O->2NaOH
0,2----------------0,4 mol
2NaOH+CO2->Na2CO3+H2O
0,4--------0,2
n Na2O=12,4\62=0,2 mol
=>C% NaOH=0,4.40\12,4+120 .100=3 %
=>m CO2=0,2.44=8,8g
\(a.CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
=> Hợp chất tạo thành làm quỳ tím hóa xanh.
b,c. Chưa đủ dữ kiện