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\(n_{Al}=0,1\left(mol\right)\\ n_{CO_2}=1,5\left(mol\right)\\ 4Al+3CO_2\underrightarrow{t^o}2Al_2O_3+3C\)
n(bd) 0,1 1,5
n(spu) 0 1,425
\(2Mg+CO_2\underrightarrow{t^o}2MgO+C\)
2,85 1,425
\(m_{hh}=2,7+24\cdot2,85=71,1\left(g\right)\\ \Rightarrow\%m_{Al}=\dfrac{2,7\cdot100\%}{71,1}\approx3,797\%\\ \Rightarrow\%m_{Mg}=100\%-3,797\%=96,203\%\)
bạn viết đề bài hơi rồi nên mk làm theo cách mk hiểu, ban kiểm tra lại thể tích CO2 xem có phải là 3,36 l khong vi mk thay so mol CO2 hoi lon
H=100%(cái này quan trọng này)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ 2Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
n(bd) 0,1 1,5
n(spu) 0 1,35\
\(2Mg+O_2\underrightarrow{t^o}2MgO\)
2,7 1,35
\(m_{hh}=2,7+24.2,7=67,5\left(g\right)\\ \%m_{Al}=\dfrac{2,7\cdot100\%}{67,5}=4\left(\%\right)\\ \Rightarrow\%m_{Mg}=100\%-4\%=96\%\)
PTHH:
2Mg + O2 =(nhiệt)=> 2MgO (1)
4Al + 3O2 =(nhiệt=> 2Al2O3 (2)
Ta có: nAl = \(\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(\Sigma n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
+) nO2 (2) = \(\dfrac{0,1.3}{4}=0,075\left(mol\right)\)
=> nO2(1) = 1,5 - 0,075 = 1,425 (mol)
=> nMg = 1,425 x 2 = 2,85 (mol)
=> mMg = 2,85 x 24 = 68,4 (gam)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{68,4}{68,4+2,7}.100\%=96,2\%\\\%m_{Al}=100\%-96,2\%=3,8\%\end{matrix}\right.\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(2Mg+O_2\underrightarrow{t^o}2MgO\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Mg}\) \(\Rightarrow n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{2,7+0,15.24}.100\%\approx42,86\%\\\%m_{Mg}\approx57,14\%\end{matrix}\right.\)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (1)
\(2Mg+O_2\underrightarrow{t^o}2MgO\) (2)
Ta có: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}\cdot\dfrac{13,5}{27}=0,375\left(mol\right)\\n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(2\right)}=0,375\left(mol\right)\) \(\Rightarrow n_{Mg}=0,75\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,75\cdot24}{0,75\cdot24+13,5}\cdot100\%\approx57,14\%\)
1)
2Mg + O2 \(\rightarrow\) 2MgO
4Al + 3O2 \(\rightarrow\)2Al2O3
Ta có: nO2=\(\frac{33,6}{22,4}\)=1,5 mol; nAl=\(\frac{2,7}{27}\)=0,1 mol
Ta có: nO2=\(\frac{1}{2}\)nMg +\(\frac{3}{4}\)nAl\(\rightarrow\) nMg=2,85 mol
\(\rightarrow\) mMg=68,4 gam
\(\rightarrow\) %Al=\(\frac{2,7}{\text{2,7+68,4}}\)=3,8%\(\rightarrow\) %Mg=96,2%