Respuesta :
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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(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:
[tex]\text{Moles of lead(II) sulfide}=\frac{265g}{293.3g/mol}=0.904mol[/tex]
The chemical equation for the reaction of lead sulfide and hydrogen peroxide follows:
[tex]PbS+4H_2O_2\rightarrow PbSO_4+4H_2O[/tex]
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 = [tex]\frac{4}{1}\times 0.904=3.616mol[/tex] 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:
[tex]3.616mol=\frac{\text{Mass of hydrogen peroxide}}{34g/mol}\\\\\text{Mass of hydrogen peroxide}=(3.616mol\times 34g/mol)=122.9g[/tex]
Hence, the mass of hydrogen peroxide needed to react completely with given amount of lead (II) sulfide is 122.9 grams.
