Question

When lead(II) sulfide reacts with oxygen (O2) gas, the products are lead(II) oxide and sulfur dioxide gas.How many grams of oxygen are required to react with 25.1 g of lead(II) sulfide?

Answer 1

8O2(g) + 8PbS(s) = 8PbO(s) + 8S(g)
rememeber that oxygen is diatomic so it exists at O2, while suphur exists as S8.
2. Use the mole ratio to find the amount of moles of oxygen so u can later use it along with the molar mass of O2(g) to find the mass.

the coeffecients on the balanced equation indicates the nuber of moles fro each substance.
Therefore comparing oxygen and Lead (ii) sulfide 8O2(g) : 8PbS(s) , they have the same ratio. Therefore, they have the same amount of moles; 8.
Now, u know that the moles of oxygen: 8 mol
The molar mass is goin to be 2(16) = 32 g/mol
use n=m/M and manipulate it so that m is by itself on one side. tHerefore, m= nM
so, m= (8mol)(32 g/mol)
m= 256 g (it makes sense because in the equation moles cancel out leaving grams)