Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (1)
\(2Mg+O_2\underrightarrow{t^o}2MgO\) (2)
Ta có: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}\cdot\dfrac{13,5}{27}=0,375\left(mol\right)\\n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(2\right)}=0,375\left(mol\right)\) \(\Rightarrow n_{Mg}=0,75\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,75\cdot24}{0,75\cdot24+13,5}\cdot100\%\approx57,14\%\)
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)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{68,4}{68,4+2,7}.100\%=96,2\%\\\%m_{Al}=100\%-96,2\%=3,8\%\end{matrix}\right.\)
\(n_{Al}=0,1\left(mol\right)\\ n_{CO_2}=1,5\left(mol\right)\\ 4Al+3CO_2\underrightarrow{t^o}2Al_2O_3+3C\)
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
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\) (1)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (2)
Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\Rightarrow n_{O_2\left(2\right)}=0,075\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(1\right)}=1,425\left(mol\right)\) \(\Rightarrow n_{Mg}=2,85\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{2,85\cdot24}{2,85\cdot24+2,7}\cdot100\%\approx96,2\%\)
\(\Rightarrow\%m_{Al}=3,8\%\)
\(n_{O_2} =\dfrac{33,6}{22,4} = 1,5(mol)\\ n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{O_2} = \dfrac{1}{2}n_{Mg} + \dfrac{3}{4}n_{Al}\\ \Rightarrow n_{Mg} = 2,85(mol)\)
Vậy :
\(\%m_{Mg} = \dfrac{2,85.24}{2,85.24 + 2,7}.100\% = 96,2\%\\ \%m_{Al} = 100\% - 96,2\% = 3,8\%\)
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: \(n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Mg}\) \(\Rightarrow n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{2,7+0,15.24}.100\%\approx42,86\%\\\%m_{Mg}\approx57,14\%\end{matrix}\right.\)