a)
\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)
\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
Ta thấy :
\(\dfrac{n_{Al}}{4} = 0,025 < \dfrac{n_{O_2}}{3} = 0,1\) nên O2 dư
Theo PTHH :
\(n_{Al_2O_3} = 0,5n_{Al} = 0,05(mol)\\ n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,075(mol)\)
Suy ra :
\(m_{Al_2O_3} = 0,05.102 = 5,1(gam)\\ m_{O_2\ dư} = (0,3 - 0,075).32 = 7,2(gam)\)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,3}{3}\) \(\Rightarrow\) Oxi còn dư, Al phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=0,05mol\\n_{O_2\left(dư\right)}=0,225mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=0,225\cdot32=7,2\left(g\right)\\m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\end{matrix}\right.\)
b) Theo PTHH: \(n_{O_2\left(pư\right)}=0,075mol\)
\(\Rightarrow V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\)
Vì Oxi chiếm khoảng 20% thể tích không khí
\(\Rightarrow V_{kk}=\dfrac{1,68}{20\%}=8,4\left(l\right)\)
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(n_{O_2}=0.3\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.1.....0.075.....0.05\)
\(m=m_{Al_2O_3}+m_{O_2\left(dư\right)}=0.05\cdot102+\left(0.3-0.075\right)\cdot32=12.3\left(g\right)\)
\(V_{kk}=5V_{O_2}=5\cdot0.075\cdot22.4=8.4\left(l\right)\)
