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H2+CuO->Cu+H2O
0,2---0,2-----0,2
Fe2O3+3H2-to>2Fe+3H2O
0,1-------0,3-------0,2
m CuO=32.\(\dfrac{50}{100}\)=16g
=>n CuO=\(\dfrac{16}{80}\)=0,2 mol
=>m Fe2O3=16g=>n Fe2O3=0,1 mol
=>m =mFe+m Cu=0,2.64+0,2.56=24g
c)Fe+H2SO4->FeSO4+H2
0,2---------------------0,2
=>m FeSO4=0,2.102=20,4g
\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ b.n_{H_2SO_4}=0,22.1,25=0,275mol\\ n_{Fe_2O_3}=a;n_{CuO}=b\\ \Rightarrow\left\{{}\begin{matrix}3a+b=0,275\\160a+80b=16\end{matrix}\right.\\ \Rightarrow a=0,075;b=0,05\\ \%m_{Fe_2O_3}=\dfrac{0,075.160}{16}\cdot100=75\%\\ \%m_{CuO}=100-75=25\%\)
\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\a, CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,05\left(MOL\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ c,V_{ddH_2SO_4}=\dfrac{0,05}{0,5}=0,1\left(l\right)\\ d,V_{ddCuSO_4}=V_{ddH_2SO_4}=0,1\left(l\right)\\ C_{MddCuSO_4}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
Gọi số mol H2O sinh ra là a (mol)
=> \(n_{H_2SO_4}=a\left(mol\right)\)
Theo ĐLBTKL: moxit + mH2SO4 = mmuối + mH2O
=> 16,6 + 98a = 24,6 + 18a
=> a = 0,1 (mol)
=> nO = 0,1 (mol)
=> mkim loại = 16,6 - 0,1.16 = 15 (g)
\(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
Ta có: \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)
a, PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
_____0,6____________________0,6 (mol)
\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)
b, Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}=\dfrac{0,6}{3}\), ta được pư hết.
Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)
Bạn tham khảo nhé!
a) 2NaOH + H2SO4 --> Na2SO4 + 2H2O
b) \(m_{NaOH}=\dfrac{200.8}{100}=16\left(g\right)\)
=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
PTHH: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,4--->0,2--------->0,2
=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
c) \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
=> \(m_{dd.H_2SO_4}=\dfrac{19,6.100}{9,8}=200\left(g\right)\)
mNaOH = 8% . 200 = 16 (g)
nNaOH = 16/40 = 0,4 (mol)
PTHH: 2NaOH + H2SO4 -> Na2SO4 + 2H2O
Mol: 0,4 ---> 0,2 ---> 0,2 ---> 0,4
mNa2SO4 = 0,2 . 119 = 23,8 (g)
mH2SO4 = 0,2 . 98 = 19,6 (g)
mddH2SO4 = 19,6/9,8% = 200 (g)
\(\left\{{}\begin{matrix}Zn\\Fe\\Mg\end{matrix}\right.+H_2SO_4\rightarrow\left\{{}\begin{matrix}ZnSO_4\\FeSO_4\\MgSO_4\end{matrix}\right.+H_2\uparrow\)
Ta có: \(m_{SO_4}=8,25-2,49=5,76\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=n_{H_2}=n_{SO_4}=\dfrac{5,76}{96}=0,06\left(mol\right)\)
a, \(m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)
b, \(V_{H_2}=0,06.22,4=1,344\)
Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
a, \(n_{H_2SO_4}=3n_{Fe_2O_3}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{9,8\%}=300\left(g\right)\)
b, \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\)
Ta có: m dd sau pư = 16 + 300 = 316 (g)
\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{316}.100\%\approx12,66\%\)
a/ Thật ra \(Fe_3O_4\) chính là hỗn hợp \(FeO;Fe_2O_3\) nên ta có được 3,48 g là khối lượng hỗn hợp \(FeO;Fe_2O_3\).
Gọi số mol của \(FeO;Fe_2O_3\) lần lược là x, y thì ta có hệ:
\(\left\{{}\begin{matrix}72x+160y=3,48\\x=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,015\\y=0,015\end{matrix}\right.\)
PTHH:
\(FeO\left(0,015\right)+H_2SO_4\left(0,015\right)\rightarrow FeSO_4\left(0,015\right)+H_2O\)
\(Fe_2O_3\left(0,015\right)+3H_2SO_4\left(0,045\right)\rightarrow Fe_2\left(SO_4\right)_3\left(0,015\right)+3H_2O\)
\(\Rightarrow n_{H_2SO_4}=0,015+0,045=0,06\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{5,88}{4,9\%}=120\left(g\right)\)
b/ Ta có: \(m_{dd}=3,48+120=123,48\left(g\right)\)
\(\left\{{}\begin{matrix}m_{FeSO_4}=0,015.152=2,28\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,015.400=6\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%\left(FeSO_4\right)=\dfrac{2,28}{123,48}.100\%=1,85\%\\C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{6}{123,48}.100\%=4,86\%\end{matrix}\right.\)