Question

Giúp mik vs

Answer 1

a) \(n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

                 0,1----------------------------->0,05

 => \(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

b) \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

Xét tỉ lệ: \(\dfrac{0,5}{4}>\dfrac{0,05}{3}\) => Al dư, O₂ hết

=> m_rắn = m_Al + m_O2 = 13,5 + 0,05.32 = 15,1 (g)