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\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)
a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
b, Ta có: \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow n_{Na}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,1.23}{0,1.23+3,9}.100\%\approx37,1\%\\\%m_K\approx62,9\%\end{matrix}\right.\)
Số mol của 2,24 lít khí H2:
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH:
\(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\left(1\right)\)
2 : 2 : 2 : 1
0,1-> 0,1 : 0,1 : 0,05(mol)
\(Na_2O+H_2O\rightarrow2NaOH\left(2\right)\)
Khối lượng của 0,2 mol Na:
\(m_{Na}=n.M=0,2.23=4,6\left(g\right)\)
Do hỗn hợp A gồm Na và H2O nên ta có:
\(m_{Hỗnhợp}=m_{Na}+m_{Na_2O}\\ \Rightarrow m_{Na_2O}=m_{Hỗnhợp}-m_{Na}\\ \Rightarrow m_{Na_2O}=12,4-4,6\\ \Rightarrow m_{Na_2O}=7,8\left(g\right)\)
1. thường gặp nhất là hỗn hợp. vd: nước tự nhiên .....
vd về chất như: nacl,......
2.
a)
\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2 (1)
0,6<----------------------0,3
=> mNa = 0,6.23 = 13,8 (g)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,1<-0,2
=> mFe = 0,1.56 = 5,6 (g)
mCu = 10 (g)
\(\left\{{}\begin{matrix}\%Na=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%Fe=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%Cu=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)
b)
PTHH: FexOy + yH2 --to--> xFe + yH2O
\(\dfrac{0,3}{y}\)<--0,3
=> \(M_{Fe_xO_y}=\dfrac{17,4}{\dfrac{0,3}{y}}=58y\left(g/mol\right)\)
=> 56x = 42y
=> \(\dfrac{x}{y}=\dfrac{3}{4}\) => CTHH: Fe3O4
a)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,6<----------------------0,3
Fe + 2HCl --> FeCl2 + H2
0,1<--0,2
=> \(\left\{{}\begin{matrix}m_{Na}=0,6.23=13,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=10\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%m_{Fe}=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%m_{Cu}=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)
b)
PTHH: FexOy + yH2 --to--> xFe + yH2O
\(\dfrac{0,3}{y}\)<--0,3
=> \(M_{Fe_xO_y}=56x+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\left(g/mol\right)\)
=> \(\dfrac{x}{y}=\dfrac{3}{4}\)
=> CTHH: Fe3O4
a)a) PTHHPTHH
2Na+2H2O−−−>2NaOH+H22Na+2H2O−−−>2NaOH+H2 (1)(1)
2Na+2C2H5OH−−−>2C2H5ONa+H22Na+2C2H5OH−−−>2C2H5ONa+H2 (2)(2)
b)b)
Gọi nH2O=a(mol),nC2H5OH=b(mol)nH2O=a(mol),nC2H5OH=b(mol)
Theo đề, ta có: 18a+46b=6,618a+46b=6,6(I)(I)
Theo PTHH, nH2=0,5a+0,5b(mol)nH2=0,5a+0,5b(mol)
nH2=2,01622,4=0,09(mol)nH2=2,01622,4=0,09(mol)
0,5a+0,5b=0,090,5a+0,5b=0,09 (II)(II)
Giai (I) và (II), {a=0,06b=0,12{a=0,06b=0,12
=>mH2O=0,06.18=1,08(g)=>mH2O=0,06.18=1,08(g)
=>mC2H5OH=6,6−1,08=5,52(g)
c)c)
VC2H5OH=5,520,8=6,9(l)VC2H5OH=5,520,8=6,9(l)
VH2O=1,081=1,08(l)VH2O=1,081=1,08(l)
=>Đr=6,9.100%6,9+1,08=86,47%