Question

Pb(CH3COO)2 + H2S → PbS + CH3COOH (Need to balance equation)

How many grams of PbS is produced when 5.00g of H2S is reacted with an excess (unlimited) supply of Pb(CH3COO)2?

Answer 1

To balance the given equation, add a coefficient of 2 in front of CH3COOH. 0.147 mol of PbS is produced, which is equal to 35.19 g of PbS.

Step-by-step explanation:

To balance the given equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's start by counting the number of atoms for each element in the given equation:

• Pb: 1 on both sides
• H: 6 on both sides
• C: 4 on both sides
• O: 2 on the reactant side and 4 on the product side
• S: 1 on the reactant side and 1 on the product side

From the above counts, we can see that the number of oxygen atoms is not balanced. We can balance the equation by adding a coefficient of 2 in front of CH3COOH:

Pb(CH3COO)2 + H2S → PbS + 2CH3COOH

Now, the equation is balanced, and we can calculate the number of grams of PbS produced. From the balanced equation, we can see that 1 mole of PbS is produced for every 1 mole of H2S reacted. The molar mass of H2S is 34.08 g/mol, so 5.00 g of H2S is equivalent to 5.00/34.08 = 0.147 mol. Therefore, 0.147 mol of PbS is produced, which is equal to 0.147 x 239.36 = 35.19 g of PbS.

Answer 2

Mass of PbS = 9.26 g

Step-by-step explanation:

Data Given :

mass of H₂S = 5.00 g

mass of PbS = ?

Reaction Given:

Pb(CH₃COO)₂ + H₂S ----→ PbS + CH₃COOH

Solution:

Balance the equation:

Pb(CH₃COO)₂ + H₂S ----→ PbS + 2CH₃COOH

Now Look for the number of moles of H₂S and PbS meta

Pb(CH₃COO)₂ + H₂S ----→ PbS + 2CH₃COOH

1 mol 1 mol

So,

1 mole of H₂S combine with excess (unlimited) supply of Pb(CH₃COO)₂ and produce 1 moles of PbS

Now Convert moles to mass for which we have to molar masses of H₂S and PbS

Molar mass of H₂S = 2 + 32 = 43 g/mol

Molar mass of PbS = 31 + 32 = 63 g/mol

Pb(CH₃COO)₂ + H₂S -------→ PbS + 2CH₃COOH

1 mol (34 g/mol) 1 mol (63 g/mol)

34 g 63 g

So,

34 g of H₂S produces 63 grams of PbS.

Now

What mass of silver is produced from 5 g of H₂S

Apply unity formula

34 g of H₂S ≅ 63 g of PbS

5 g of H₂S ≅ X g of PbS

By doing cross multiplication

Mass of PbS = 63 g x 5 g / 34 g

Mass of PbS = 9.26 g