Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
nH2 = 2.24/22.4 = 0.1 (mol)
Na + H2O => NaOH + 1/2 H2
0.2....................0.2..........0.1
mNa = 0.2 * 23 = 4.6 (g)
mNa2O = 17 - 4.6 = 12.4 (g)
nNa2O = 12.4/62 = 0.2 (mol)
Na2O + H2O => 2NaOH
0.2........................0.4
nNaOH = 0.2 + 0.4 = 0.6 (mol)
mNaOH = 0.6 * 40 = 24 (g)
nCuO = 24/80 = 0.3 (mol)
CuO + H2 -t0-> Cu + H2O
1...........1
0.3.........0.1
LTL : 0.3/1 > 0.1/1
=> CuO dư
nCu = nH2 = 0.1 (mol)
mCu = 0.1 * 64 = 6.4 (g)
\(a,m_{rắn}=m_{Cu}=2,7\left(g\right)\\ \Rightarrow m_{\left(Zn,Fe\right)}=12-2,7=9,3\left(g\right)\\ n_{H_2}=0,15\left(mol\right),n_{axit}=2.0,2=0,4\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,15}{1}< \dfrac{0,4}{1}\Rightarrow axit.dư\\ \Rightarrow\left\{{}\begin{matrix}65+56b=9,3\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2,7}{12}.100=22,5\%\\ \%m_{Zn}=\dfrac{0,1.65}{12}.100\approx54,167\%\\ \%m_{Fe}=\dfrac{0,05.56}{12}.100\approx23,333\%\)
\(b,ddA:FeCl_2,ZnCl_2,H_2SO_4\left(dư\right)\\ m_{ddH_2SO_4}=200.1,14=228\left(g\right)\\ m_{ddA}=m_{\left(Zn,Fe\right)}+m_{ddH_2SO_4}-m_{H_2}=9,3+228-0,15.2=237\left(g\right)\)
\(C\%_{ddZnCl_2}=\dfrac{136.0,1}{237}.100\approx5,738\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,05}{237}.100\approx2,679\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,15\right).98}{237}.100\approx10,338\%\)
Đã sửa lần cuối lúc 20:45
a) Cu + 2H2SO4 → CuSO4 + SO2↑ + 2H2O
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{SO_2}=n_{Cu}=0,2\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
\(\%m_{Cu}=\dfrac{12,8}{20,8}.100=61,54\%\); \(\%m_{CuO}=38,46\%\)
b) \(n_{CuO}=\dfrac{20,8-12,8}{80}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=0,2.2+0,1=0,5\left(mol\right)\)
\(m_{ddH_2SO_4}=\dfrac{0,5.98}{80\%}=61,25\left(g\right)\)
\(n_{CuSO_4}=0,2+0,1=0,3\left(mol\right)\)
\(m_{CuSO_4}=0,3.160=48\left(g\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Cu}:x\left(mol\right)\\n_{CuO}:y\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}m_{H2SO4}=\frac{70.112}{100}=78,4\left(g\right)\\n_{H2SO4}=\frac{78,4}{98}=0,8\left(mol\right)\end{matrix}\right.\)
\(Cu+2H_2SO_4\rightarrow CuSO_4+2H_2O+SO_2\)
x___2x___________x_________2x_______x
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
y________y_______y___________y
Giải hệ PT:
\(\left\{{}\begin{matrix}64x+80y=28\\2x+y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,375.65=24\left(g\right)\\m_{CuO}=0,05.80=4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\frac{24}{28}.100\%=85,71\%\\\%m_{CuO}=100\%-85,71\%=14,29\%\end{matrix}\right.\)
goi x la so mol cua Cu
y la so mol cua CuO
mH2SO4=70.112\100=78,4g
nH2SO4=78,4\98=0,8(mol)
Cu+2H2SO4(d,n)to→to→CuSO2+2H2O+SO2
de: x 2x x 2x x
CuO + H2SO4→→ CuSO4 +H2O
de: y y y y
Ta co: 64x + 80y = 28
2x + y = 0,8
⇒{x=0,375(mol)y=0,05(mol)
mCu=0,375.64=24g
mCuO=0,05.80=4g
%mCu=24\28.100%≈85,71%%
%mCuO=4\28.100%≈14,29%