Question

đốt m g Al trong O2 thu đc 5,1g Al2O3, m=?

 

Answer 1

PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Ta có: \(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)

\(\Rightarrow n_{Al}=0,1mol\) \(\Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\)

 

Answer 2

_{\(4Al+3O_2\rightarrow2Al_20_3\)}

\(n_{Al_2O_3}=\dfrac{5.1}{102}=0.05\left(mol\right)\)

theo phương trình:

\(n_{Al}=n_{Al_2O_3}=0.05\left(mol\right)\)

\(m_{Al}=n_{Al}\times M_{Al}=0.05\times27=1.35\left(g\right)\)

Answer 3