Question

Answer 1

\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

0,1    0,075           0,05           ( mol )

\(m_{Al_2O_3}=0,05.102=5,1g\)

\(V_{O_2}=0,075.22,4=1,68l\)

\(V_{kk}=V_{O_2}.5=1,68.5=8,4l\)

Answer 2

4Al+3O2-to>2Al2O3

0,1----0,075----0,05

n Al=\(\dfrac{2,7}{27}\)=0,1 mol

=>Vkk=0,075.22,4.5=8,4l

=>m Al2O3=0,05.102=5,1g

Answer 3

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ Theo.pt:n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ m_{Al_2O_3}=0,05.102=5,1\left(g\right)\\ Theo.pt:n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.0,1=0,075\left(mol\right)\\ V_{kk}=5.0,075.22,4=8,4\left(l\right)\)