Question

Lead (II) sulfide, PbS, reacts with oxygen gas to produce lead (II) oxide and sulfur dioxide. If 0.750 moles of O2 were used during this chemical reaction, how many grams of lead (II) oxide would be produced?

Answer 1

Answer:

111.5 grams of lead (II) oxide would be produced.

Explanation:

$2PbS+3O_2\rightarrow 2PbO+2SO_2$

Moles of oxygen = 0.750 mol

According to reaction, 3 moles of oxygen gas gives 2 moles of lead(II) oxide ,then 0.750 moles of oxygen gas will give:

$\frac{2}{3}\times 0.750 mol=0.50 mol$ of lead (II) oxide

Mass of 0.50 moles lead(II) oxide:

0.50 mol × 223 g/mol = 111.5 g

111.5 grams of lead (II) oxide would be produced.