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PTHH: 2Na + 2H2O --> H2 + 2NaOH
Na2O + H2O -> 2NaOH
Ta có: nH2= \(\dfrac{3,36}{22,4}\) = 0,15(mol)
Theo pthh(1) ta có: nNa = 2nH2= 0,3(mol)
=> mNa = 0,323=6,9(g)
ta có: mNa2O= 13,1-6,9= 6,2(g)
\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)
$n_{H_2} = \dfrac{1,792}{22,4} = 0,08(mol)$
$n_{HCl} = 0,2(mol)$
$2Na + 2H_2O \to 2NaOH + H_2$
$Na_2O + H_2O \to 2NaOH$
$NaOH + HCl \to NaCl + H_2O$
Theo PTHH :
$n_{Na} = 2n_{H_2} = 0,16(mol)$
$2n_{Na_2O} + n_{Na} = n_{NaOH} = n_{HCl} = 0,2$
$\Rightarrow n_{Na_2O} = 0,02(mol)$
$\%m_{Na} = \dfrac{0,16.23}{5}.100\% = 73,6\%$
$\%m_{Na_2O} = \dfrac{0,02.62}{5}.100\% = 24,8\%$
Do HNO3 nóng dư nên Fe, Cu pứ hết --> Fe3+ & Cu2+
M(B) = 36 --> nNO : nNO2 = 5:3
Khi cho đ sau pứ tác dụng vs NH3 dư thì --> Fe(OH)3 ko tan, Cu(NH3)4(OH)2 tan
--> Chất rắn sau nung: Fe2O3: n = 0,05 --> nFe = 0,1 -->mFe = 5,6, mCu = 6,4g
Từ nFe, nCu, bảo toàn electron --> nNO, nNO2 --> V
c, Dung dịch kiềm> Vì trong dd D có NH4NHO3, nên cho kiềm vào sẽ sinh ra NH3.
2Na+2H2O->2NaOH+H2
0,5-----0,5-----------0,5----0,25
Na2O+H2O->2NaOH
0,1--------0,1-----------0,2
n H2=0,25 mol
=>m Na =0,5.23=11,5g
=>m Na2O=6,2g=>n Na2O=0,1 mol
=>m NaOH=0,7.40=28g
=>VH2O=0,6.22,4=13,44l
a)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: x 1,5x
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)
b)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,1 0,15 0,05
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,1 0,1 0,1
\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)
mdd sau pứ = 5,1+250-0,15.2 = 254,8(g)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)
\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)
a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Na}=2n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Na}=0,3.23=6,9\left(g\right)\)
\(\Rightarrow m_{Na_2O}=13,1-6,9=6,2\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,5\left(mol\right)\)
Ta có: m dd sau pư = 13,1 + 200 - 0,15.2 = 212,8 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,5.40}{212,8}.100\%\approx9,4\%\)