How many grams of oxygen are required to react with 0.128 lead(II) sulfide?
How many grams of oxygen are required to react with 31.5 g of lead(II) sulfide?
To determine the number of grams of oxygen required to react with 0.128 grams of lead(II) sulfide (PbS), we need to use the balanced chemical equation for the reaction between lead(II) sulfide and oxygen.
The balanced equation for the reaction is:
PbS + O2 -> PbO + SO2
From the equation, we can see that 1 mole of PbS reacts with 1 mole of O2.
To determine the number of moles of PbS in 0.128 grams, we need to divide the mass of PbS by its molar mass.
The molar mass of PbS is calculated as follows:
Molar mass of Pb: 207.2 g/mol
Molar mass of S: 32.1 g/mol
Molar mass of PbS = Molar mass of Pb + Molar mass of S
= 207.2 g/mol + 32.1 g/mol
= 239.3 g/mol
Now, let's calculate the number of moles of PbS:
Number of moles = Mass of PbS / Molar mass of PbS
= 0.128 g / 239.3 g/mol
≈ 0.000535 moles
Since the reaction is 1:1, the same amount of moles of O2 is required.
Therefore, the number of moles of O2 needed = 0.000535 moles.
To calculate the mass of O2 in grams, we need to use the molar mass of O2, which is 32.0 g/mol.
Mass of O2 = Number of moles of O2 × Molar mass of O2
= 0.000535 moles × 32.0 g/mol
≈ 0.0171 grams
Hence, approximately 0.0171 grams of oxygen are required to react with 0.128 grams of lead(II) sulfide.
To determine the number of grams of oxygen required to react with 0.128 grams of lead(II) sulfide (PbS), we need to use balanced chemical equations and stoichiometry.
First, let's write the balanced equation for the reaction between lead(II) sulfide and oxygen:
2PbS + 3O2 -> 2PbO + 2SO2
From the balanced equation, we can see that it takes 3 moles of oxygen (O2) to react with 2 moles of lead(II) sulfide (PbS).
To find the number of moles of lead(II) sulfide, we can use the formula:
moles = mass / molar mass
The molar mass of lead(II) sulfide (PbS) is the sum of the molar masses of lead (Pb) and sulfur (S). The atomic masses of lead and sulfur are 207.2 g/mol and 32.1 g/mol, respectively.
molar mass of PbS = (207.2 g/mol) + (32.1 g/mol) = 239.3 g/mol
Now, let's calculate the number of moles of lead(II) sulfide:
moles of PbS = 0.128 g / 239.3 g/mol ≈ 0.0005356 mol
From the balanced equation, we know that 3 moles of oxygen react with 2 moles of lead(II) sulfide. Therefore, we can set up a proportion:
(3 moles of O2) / (2 moles of PbS) = (x moles of O2) / (0.0005356 moles of PbS)
Solving for x:
x = (3 moles of O2) × (0.0005356 moles of PbS) / (2 moles of PbS) ≈ 0.00203 moles of O2
Finally, let's convert the moles of oxygen to grams using the molar mass of oxygen (O2), which is 32 g/mol:
grams of O2 = (0.00203 moles of O2) × (32 g/mol) ≈ 0.065 g of O2
Therefore, approximately 0.065 grams of oxygen are required to react with 0.128 grams of lead(II) sulfide.