How many grams of lead ii sulfide is produced when 25.0 g lead ii acetate reacts with excess hydrogen sulfide

18.42 grams of lead (II) sulfide will be produced in the given reaction:

Explanation: The reaction of lead (II) acetate and hydrogen sulfide follows:

To calculate the moles, we use the formula:

   ....(1)

Molar mass of lead (II) acetate = 325.29 g/mol

Given mass of lead (II) acetate = 25 g

Putting values in above equation, we get:

We are given that hydrogen sulfide is present in excess, so limiting reagent is lead (II) acetate because it limits the formation of product.

By stoichiometry of the reaction,

1 moles of lead (II) acetate produces 1 mole of lead (II) sulfide

So, 0.0768 moles of lead (II) acetate will produce = = 0.0768 moles of lead (II) sulfide.

Now, to calculate the mass of lead (II) sulfide, we use equation 1, we get:

Molar mass of lead (II) sulfide = 239.3 g/mol