Respuesta :
Answer: 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:
[tex](CH_3COO)_2Pb+H_2S\rightarrow PbS+2CH_3COOH[/tex]
To calculate the moles, we use the formula:
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(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:
[tex]Moles=\frac{25g}{325.29g/mol}=0.0768moles[/tex]
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 = [tex]\frac{1}{1}\times 0.0768[/tex] = 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
[tex]0.0768mol=\frac{\text{Given mass}}{239.3g/mol}[/tex]
[tex]\text{Mass of lead (II) sulfide}=18.42g[/tex]
