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PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\) (1)
\(Na_2O+H_2O\rightarrow2NaOH\) (2)
a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Na}=0,2\left(mol\right)\)
\(\Rightarrow m_{Na}=0,2\cdot23=4,6\left(g\right)\) \(\Rightarrow m_{Na_2O}=6,2\left(g\right)\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{NaOH\left(1\right)}=n_{Na}=0,2\left(mol\right)\\n_{NaOH\left(2\right)}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{NaOH}=0,4\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,4\cdot40=16\left(g\right)\)
c) Theo các PTHH: \(n_{H_2O}=n_{Na}+n_{Na_2O}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,3\cdot18=5,4\left(g\right)\)
a) Gọi số mol Na, Ca là a, b (mol)
=> 23a + 40b = 8,3 (1)
PTHH: 2Na + 2H2O --> 2NaOH + H2
a--------------->a------>0,5a
Ca + 2H2O --> Ca(OH)2 + H2
b--------------->b--------->b
=> \(n_{H_2}=0,5a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) (2)
(1)(2) => a = 0,1; b = 0,15
=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,1.23}{8,3}.100\%=27,71\%\\\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\%=72,29\%\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m_{NaOH}=0,1.40=4\left(g\right)\\m_{Ca\left(OH\right)_2}=0,15.74=11,1\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ca}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+40b=8,3\\0,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ b,\Rightarrow\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\approx72,289\%\\ \Rightarrow\%m_{Na}\approx27,711\%\\ b,n_{NaOH}=a=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=b=0,15\left(mol\right)\\ m_{bazo}=m_{NaOH}+m_{Ca\left(OH\right)_2}=40.0,1+74.0,15=15,1\left(g\right)\)
2Na+2H2O->2NaOH+H2
x------------------------------0,5x
Ca+2H2O->Ca(OH)2+H2
y-------------------------------y
Ta có :
\(\left\{{}\begin{matrix}23x+40y=8,3\\0,5x+y=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
=>%mNa=\(\dfrac{0,1.23}{8,3}.100=27,71\%\)
=>%mCa=72,29%
b)m bazo=0,1.40+0,15.74=15,1g
\(a) Ca + 2H_2O \to Ca(OH)_2 + H_2\\ CaO + H_2O \to Ca(OH)_2\\ n_{Ca} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ m_{Ca} = 0,15.40 = 6(gam)\\ \Rightarrow m_{CaO} = 17,2 - 6 = 11,2(gam)\\ b) n_{Ca(OH)_2} = n_{Ca} + n_{CaO} = 0,15 + \dfrac{11,2}{56} = 0,35(mol)\\ m_{Ca(OH)_2} = 0,35.74 = 25,9(gam)\)
a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ 2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)
Cho hỗn hợp tác dụng với NaOH, chất rắn không tan là Fe
=> mFe= 1,12 (g) \(\Rightarrow n_{Fe}=0,02\left(mol\right)\)
Ta có: \(n_{H_2\left(2\right)}=n_{Fe}=0,02\left(mol\right)\)
=> \(n_{H_2\left(1\right)}=\Sigma n_{H_2}-n_{H_2\left(2\right)}=0,065-0,02=0,045\left(mol\right)\)
\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2\left(1\right)}=0,03\left(mol\right)\)
\(\Rightarrow m_{Al}=0,03.27=0,81\left(g\right)\)
\(\Rightarrow\%m_{Al}=41,97\%,\%m_{Fe}=58,03\%\)
b) \(m_{FeCl_2}=0,02.127=2,54\left(g\right)\\ m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)
2Na+2H2O--->2NaOH+H2
x----------------------------0,5x
2K+2H2O---->2KOH+H2
y--------------------------0,5y
a) n\(_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo bài ra ta có
\(\left\{{}\begin{matrix}23x+39y=6,2\\0,5x+0,5y=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
m\(_{Na}=0,1.23=2,3\left(g\right)\)
m\(_K=0,1.39=3,9\left(g\right)\)
b) Theo pthh
n\(_{kiềm}=n_{KL}=0,2\left(mol\right)\)
m\(_{kiềm}=0,2\left(40+56\right)=19,2\left(g\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\) (1)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\uparrow\) (2)
a) Đặt \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_{Ca}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow23a+40b=1,6\) (*)
Ta có: \(n_{H_2\left(1\right)}+n_{H_2\left(2\right)}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow\dfrac{1}{2}a+b=0,25\) (**)
Từ (*) và (**) \(\Rightarrow\) Hệ có nghiệm âm
*Bạn xem lại đề !!