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a/ \(2Mg\left(0,02\right)+O_2\left(0,01\right)\rightarrow2MgO\) (1)
\(3Fe\left(0,03\right)+2O_2\left(0,02\right)\rightarrow Fe_3O_4\)(2)
\(n_{Mg}=\dfrac{0,48}{24}=0,02\left(mol\right)\)
\(n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
\(\Rightarrow n_{O_2\left(2\right)}=0,03-0,01=0,02\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,03.56=1,68\left(g\right)\)
\(\Rightarrow m_{hh}=1,68+0,48=2,16\left(g\right)\)
b/ \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,48}{2,16}.100\%=22,22\%\\\%Fe=100\%-22,22\%=77,78\%\end{matrix}\right.\)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (1)
\(2Mg+O_2\underrightarrow{t^o}2MgO\) (2)
Ta có: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}\cdot\dfrac{13,5}{27}=0,375\left(mol\right)\\n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(2\right)}=0,375\left(mol\right)\) \(\Rightarrow n_{Mg}=0,75\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,75\cdot24}{0,75\cdot24+13,5}\cdot100\%\approx57,14\%\)
\(n_{Mg}=\dfrac{0.48}{24}=0.02\left(mol\right)\)
\(n_{O_2}=\dfrac{0.672}{22.4}=0.03\left(mol\right)\)
\(2Mg+O_2\underrightarrow{t^0}2MgO\)
\(0.02...0.01\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.03...\left(0.03-0.01\right)\)
\(m_{Fe}=0.03\cdot56=1.68\left(g\right)\)
\(m_{hh}=1.68+0.48=2.16\left(g\right)\)
\(\left\{{}\begin{matrix}Mg:a\left(mol\right)\\Al:b\left(mol\right)\\Fe:c\left(mol\right)\end{matrix}\right.\)
Theo đề bài , ta có :
\(\dfrac{24a+10}{24a+27b+56c}.100\%=28,57\%\)
\(\dfrac{24a}{24a+27b+56c-25}.100\%=14,286\%\)
Gọi : 24a = x ; 24a+27b+56c = y
Ta có :
\(\dfrac{x+10}{y}=\dfrac{28,57}{100}\) ; \(\dfrac{x}{y-25}=\dfrac{14,286}{100}\)
Suy ra: x = 2,8579; y = 45
Suy ra :
\(\%m_{Mg}=\dfrac{2,8579}{45}.100\%=6,35\%\)
mhỗn hợp = y = 45(gam)
a. PTHH:
Fe + 2HCl ---> FeCl2 + H2 (1)
Mg + 2HCl ---> MgCl2 + H2 (2)
b. Gọi x, y lần lượt là số mol của Fe và Mg
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT(1): \(n_{H_2}=n_{Fe}=x\left(mol\right)\)
Theo PT(2): \(n_{H_2}=n_{Mg}=y\left(mol\right)\)
\(\Rightarrow x+y=0,25\) (*)
Theo đề, ta lại có: 56x + 24y = 8,25 (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}x+y=0,25\\56x+24y=8,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,07\\y\approx0,18\end{matrix}\right.\)
=> \(m_{Fe}=0,07.56=3,92\left(g\right)\)
=> \(\%_{m_{Fe}}=\dfrac{3,92}{8,25}.100\%=47,52\%\)
\(\%_{m_{Mg}}=100\%-47,52\%=52,48\%\)
a,đổi.672ml=0,672l
tổng.số.mol.O2.cần.dùng.là:n=V/22,4=0,672/22,4=0,03(mol)
(1)3Fe+2O2--->Fe3O4
(2)2Mg+O2--->2MgO
số.mol.Mg.dùng.trong.PƯ.là:n=m/M=0,48/24=0,02(mol)
theo.PT(.2,)số.mol.O2.cần.dùng.để.tham.gia.PƯ.là:0,02x1:2=0,01(mol)
=>số.mol.O2.dùng.cho.PƯ.(1).là:0,03-0,01=0,02(mol)
theo.PT.(1),số.mol.Fe.là:0,02x3:2=0,03(mol)
Khối.lượng.sắt.là:0,03x56=1,68(g)
khối.lượng.hỗn.hợp.ban.đầu.là.:0,48+1,68=2,16(g)
b,%Fe=1,68x100/2,16=77,(7)%
=>%Mg=100%-77,(7)%=22,(2)%