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Ta có nH2 = 0,8,9622,4\" id=\"MathJax-Element-1-Frame\">896\\22,4= 0,04 ( mol )
\n\n2Al + 3H2SO4 →\" id=\"MathJax-Element-2-Frame\">→→ Al2(SO4)3 + 3H2
\n\nx............1,5x............x...................1,5x
\n\nFe + H2SO4 →\" id=\"MathJax-Element-3-Frame\">→→ FeSO4 + H2
\n\ny.........y.............y.................y
\n\n=> {27x+56y=111,5x+y=0,4\" id=\"MathJax-Element-4-Frame\">{27x+56y=1,1
\n\n{27x+56y=111,5x+y=0,4\">.......1,5x+y=0,04
\n\n=> {x=0,2y=0,1\" id=\"MathJax-Element-5-Frame\">{x=0,02y=0,01
\n\n=> mAl = 0,02 . 27 = 0,54 ( gam )
\n\n=> %mAl =0, 5,411×100≈49,1%\" id=\"MathJax-Element-6-Frame\">54\\1,1×100≈49,1%
\n\n=> %mFe = 100 - 49,1 = 50,9 ( %)
\n\n=> nH2SO4 = 1,5 . 0,02 + 0,01 = 0,04 ( mol )
\n\nmình đang nghĩ >>
\n
Câu 1:
\(Zn\rightarrow Zn^{+2}+2e\)
\(2H^++2e\rightarrow H_2^o\)
Ta có:
\(m_{Zn}=\frac{9,45}{65}=0,15\left(mol\right)\)
Bảo toàn e:
\(2n_{Zn}=2n_{H2}\Rightarrow n_{H2}=n_{Zn}=0,15\left(mol\right)\)
Câu 2:
\(Al\rightarrow Al^{+3}+3e\)
\(2H^++2e\rightarrow H_2^o\)
Ta có:
\(n_{H2}=\frac{10,80}{22,4}=0,45\left(mol\right)\)
Bảo toàn e:
\(3n_{Al}=2n_{H2}\)
\(\Rightarrow n_{Al}=\frac{2}{3}n_{H2}=0,3\left(mol\right)\)
Câu 3:
\(Cu\rightarrow Cu^++2e\)
\(S^{+6}+2e\rightarrow S^{+4}\)
Ta có:
\(n_{Cu}=\frac{19,2}{64}=0,3\left(mol\right)\)
Bảo toàn e:
\(2n_{Cu}=2n_{SO2}\Rightarrow n_{SO2}=n_{Cu}=0,3\left(mol\right)\)
Câu 4:
\(Zn\rightarrow Zn^{+2}+2e\)
\(Al\rightarrow Al^{+3}+3e\)
\(O_2+4e\rightarrow2O^{-2}\)
Gọi \(\left\{{}\begin{matrix}n_{Zn}:x\left(mol\right)\\n_{Al}:y\left(mol\right)\end{matrix}\right.\)
Bảo toàn e:
\(2n_{Zn}+3n_{Al}=4n_{O2}\)
\(2x+3y=1\Leftrightarrow x=y=0,2\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=13\left(g\right)\\m_{Al}=5,4\left(g\right)\end{matrix}\right.\)
Câu 5:
\(Mg\rightarrow Mg^{+2}+2e\)
\(Fe\rightarrow Fe^{+3}+3e\)
\(S^{+6}+2e\rightarrow S^{+4}\)
Bảo toàn e:
\(2n_{Mg}+3n_{Fe}=2n_{SO2}\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a_____a_________________ a
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b_____ b ___________________b
Giải hệ PT:
\(\left\{{}\begin{matrix}56a+65b=12,1\\a+b=\frac{4,48}{22,4}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=100\%-46,28\%=53,72\%\end{matrix}\right.\)
\(\Rightarrow V_{H2SO4\left(can.dung\right)}=\frac{0,1+0,1}{0,2}=0,1\left(l\right)\)
\(\Rightarrow\left\{{}\begin{matrix}CM_{FeSO4}=\frac{0,1}{0,1}=1M\\CM_{ZnSO4}=\frac{0,1}{0,1}=1M\end{matrix}\right.\)
a. nSO2 = 0,7 mol
2Fe + 6H2SO4 ➝ Fe2(SO4)3 + 3SO2 + 6H2O
Cu + 2H2SO4 ➝ CuSO4 + SO2 + 2H2O
Gọi số mol của Fe và Cu lần lượt là x, y ta có hệ:
\(\left\{{}\begin{matrix}56x+64y=40\\\frac{3}{2}x+y=0,7\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=0,12\\y=0,52\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Fe}=6,72g\\m_{Cu}=33,28g\end{matrix}\right.\)
b. nH2SO4pứ = 0,36 + 1,04 = 1,4 mol => m = 137,2 g
c. mmuối = mFe2(SO4)3 + mCuSO4 = 107,2 g
Ta có:
\(n_{SO2}=\frac{2,668}{22,4}=0,12\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Zn}:x\left(mol\right)\\n_{ZnO}:y\left(mol\right)\end{matrix}\right.\)
\(Zn+2H_2SO_4\rightarrow ZnSO_4+SO_2+2H_2O\)
x_____2x__________________x______
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
y_______y____________________
\(\Rightarrow x=0,12\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,12.65=7,8\left(g\right)\\m_{ZnO}=27,24-7,8=19,44\left(g\right)\end{matrix}\right.\)
\(n_{ZnO}=\frac{19,44}{81}=0,24\left(mol\right)\)
\(n_{H2SO4}=0,48\left(mol\right)\)
\(\Rightarrow m_{H2SO4}=0,48.96=46,08\left(g\right);m_{dd\left(H2SO4\right)}=\frac{46,08}{80\%}=57,6\left(g\right)\)