Question

Answer 1

Bài 3:

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: \(2KMnO_4\xrightarrow[t^o]{}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,5\left(mol\right)\)

=> \(m_{KMnO_4}=0,5.158=79\left(g\right)\)

Bài 4:

\(n_{HgO}=\dfrac{81,375}{217}=0,375\left(mol\right)\)

PTHH: \(2HgO\xrightarrow[]{t^o}2Hg+O_2\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{HgO}=0,1875\left(mol\right)\)

=> \(V_{O_2}=0,1875.22,4=4,2\left(l\right)\)