Question

In an experiment to determine the formula of an oxide of lead metal,excess hydrogen gas was passed over 100g of the heated oxide 8.67g lead metal was formed.calculate the formula of the oxide

Answer 1

Remark:

Apparently the question contains a typo, and the correct mass of oxide is 10.0 g

So, the question:

In an experiment to determine the formula of an oxide of lead metal,excess hydrogen gas was passed over 10.0g of the heated oxide 8.67g lead metal was formed.calculate the formula of the oxide

Solution:

Pb_xO_y+yH₂ $\to$ xPb+yH₂O - general formula of reaction

m_o=m_PbxOy-m_Pb=10.0g-8.67g=1.33g

n_o=m_o/A_o=1.33g/(16g*mol^-1)=0.083mol

n_Pb=m_Pb/A_Pb=8.67g/(207g*mol^-1)=0.042mol

x/y=n_Pb/n_o=0.042mol/0.083mol =1/2

So, the formula of oxide is PbO₂

Answer: PbO₂