Cho 8,3g hỗn hợp A gồm(Na và Ca) tác dụng với H2O thi đc 4,48l H2(đktc)
a.Tính% khối lượng mỗi kim loại trong hỗn hợp A
b.Tính khối lượng bazơ thu đc sau pư
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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) \(\left\{{}\begin{matrix}m_{NaOH}=0,1.40=4\left(g\right)\\m_{Ca\left(OH\right)_2}=0,15.74=11,1\left(g\right)\end{matrix}\right.\)
a)
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Theo PTHH : $n_{Zn} = n_{H_2} = \dfrac{2,8}{22,4} = 0,125(mol)$
$m_{Zn} = 0,125.65 = 8,125(gam)$
$m_{Cu} = 8,3 - 8,125 = 0,175(gam)$
$\%m_{Zn} = \dfrac{8,125}{8,3}.100\% = 97,9\%$
$\%m_{Cu} = 100\% -97,9\% = 2,1\%$
$n_{H_2SO_4} = n_{H_2} = 0,125(mol) \Rightarrow m_{H_2SO_4} = 0,125.98 = 12,25(gam)$
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
\(n_{H_2}=\frac{3,92}{22,4}=0,175mol\)
a. PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
b. Đặt \(\hept{\begin{cases}x\left(mol\right)=n_{Mg}\\y\left(mol\right)=n_{Al}\end{cases}}\)
\(\rightarrow24x+27y=3,75\left(1\right)\)
Theo phương trình \(n_{Mg}+1,5n_{Al}=n_{H_2}=0,175\)
\(\rightarrow x+1,5y=0,175\left(2\right)\)
Từ (1) và (2) \(\rightarrow\hept{\begin{cases}x=0,1mol\\y=0,05mol\end{cases}}\)
\(\rightarrow m_{Mg}=0,1.24=2,4g\)
\(\rightarrow m_{Al}=3,75-2,4=1,35g\)
\(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\)
a(mol)...\(\dfrac{3a}{2}\left(mol\right)\)................................\(\dfrac{3a}{2}\left(mol\right)\)
\(Fe+H_2SO_4-->FeSO_4+H_2\)
b(mol)....b(mol)..........................b(mol)
a) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo bài ra ta có hệ:
\(\left\{{}\begin{matrix}27a+56b=8,3\\\dfrac{3a}{2}+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
=> \(m_{Al}=0,1.27=2,7\left(g\right)\)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
b) \(\%m_{Al}=\dfrac{2,7}{8,3}.100\%\approx32,53\%\)
\(\%m_{Fe}=100\%-32,53\%\approx67,47\%\)
c) \(m_{ct\left(H_2SO_4\right)}=\left(\dfrac{3}{2}.0,1+0,1\right).98=24,5\left(g\right)\)
=> \(m_{dd\left(H_2SO_4\right)}=\dfrac{24,5.100}{4,9}=500\left(ml\right)\)
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!
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ca}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+40b=8,3\\0,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ b,\Rightarrow\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\approx72,289\%\\ \Rightarrow\%m_{Na}\approx27,711\%\\ b,n_{NaOH}=a=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=b=0,15\left(mol\right)\\ m_{bazo}=m_{NaOH}+m_{Ca\left(OH\right)_2}=40.0,1+74.0,15=15,1\left(g\right)\)
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