a)
\(Al + 4HNO_3 \to Al(NO_3)_3 + NO + 2H_2O\)
b)
\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\)
Theo PTHH : \(n_{NO} = n_{Al} = 0,1(mol)\)
\(\Rightarrow V = 0,1.22,4 = 2,24(lít)\)
a,\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)
b,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
Theo PTHH\(n_{NO}=n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow=0,1.22,4=2,24\left(l\right)\)