Question

Lead(II) sulfide was once used in glazing earthenware. It will also react with hydrogen peroxide to form lead(II) sulfate and water. How many grams of hydrogen peroxide are needed to react completely with 265 g of lead(II) sulfide?

Answer 1

Answer: The mass of hydrogen peroxide needed to react completely with given amount of lead (II) sulfide is 122.9 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of lead(II) sulfide = 265 g

Molar mass of lead(II) sulfide = 293.3 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead(II) sulfide}=\frac{265g}{293.3g/mol}=0.904mol$

The chemical equation for the reaction of lead sulfide and hydrogen peroxide follows:

$PbS+4H_2O_2\rightarrow PbSO_4+4H_2O$

By Stoichiometry of the reaction:

1 mole of lead(II) sulfide reacts with 4 moles of hydrogen peroxide

So, 0.904 moles of lead(II) sulfide will react with = $\frac{4}{1}\times 0.904=3.616mol$ of hydrogen peroxide

Now, calculating the mass of hydrogen peroxide from equation 1, we get:

Molar mass of hydrogen peroxide = 34 g/mol

Moles of hydrogen peroxide = 3.616 moles

Putting values in equation 1, we get:

$3.616mol=\frac{\text{Mass of hydrogen peroxide}}{34g/mol}\\\\\text{Mass of hydrogen peroxide}=(3.616mol\times 34g/mol)=122.9g$

Hence, the mass of hydrogen peroxide needed to react completely with given amount of lead (II) sulfide is 122.9 grams.