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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) \(\left\{{}\begin{matrix}m_{NaOH}=0,1.40=4\left(g\right)\\m_{Ca\left(OH\right)_2}=0,15.74=11,1\left(g\right)\end{matrix}\right.\)
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: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b: Đặt \(a=n_{Zn};b=n_{Fe}\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}65a+56b=37.2\\a+b=0.6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.4\\b=0.2\end{matrix}\right.\)
\(m_{Zn}=0.4\cdot65=25\left(g\right)\)
\(m_{Fe}=0.5\cdot56=11.2\left(g\right)\)
2Na+2H2O->2NaOH+H2
x-------------------x----------0,5x mol
Ba+2H2O->Ba(OH)2+H2
y---------------------y----------y mol
aTa có :)\(\left\{{}\begin{matrix}23x+137y=2,06\\0,5x+y=0,025\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,03\\y=0,01\end{matrix}\right.\)
=>mbazo=0,03.40+0,01.171=2,91g
=>m Na=0,03.23=0,69g
=>m Ba=0,01.137=1,27g
a, Cu không tác dụng với dd HCl.
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
Bạn xem lại xem đề cho bao nhiêu gam hỗn hợp nhé, vì mZn đã bằng 13 (g) rồi.
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Ag}=20-13=7\left(g\right)\)
b, \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{20}.100\%=65\%\\\%m_{Ag}=100-65=35\%\end{matrix}\right.\)
2Al+6HCl->2AlCl3+3H2
x-----------------x---------\(\dfrac{3}{2}\)x
Zn+2HCl->ZnCl2+H2
y---------------y--------y
Ta có :
\(\left\{{}\begin{matrix}27x+65y=24,9\\\dfrac{3}{2}x+y=0,6\end{matrix}\right.\)
=>x=0,2 mol ,y=0,3 mol
=>m AlCl3= 0,2.133,5=26,7g
=>m ZnCl2 =0,3.136=40,8g
=>%mAl=\(\dfrac{0,2.27}{24,9}.100\)=21,69%
=>%m Zn=78,31%
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ca}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+40b=8,3\\0,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ b,\Rightarrow\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\approx72,289\%\\ \Rightarrow\%m_{Na}\approx27,711\%\\ b,n_{NaOH}=a=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=b=0,15\left(mol\right)\\ m_{bazo}=m_{NaOH}+m_{Ca\left(OH\right)_2}=40.0,1+74.0,15=15,1\left(g\right)\)
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