Balanced Equation:

2PbS+ O2 --> 2PbO +2SO2

How many grams of oxygen are required to react with 0.124mole lead(II) sulfide?

Ummmm.. I tried to convert it to grams?
0.124 mol * molar mass of PbS (239.3) = 29.67 g of PbS then tried to plug it into the equation..
1/2 * 29.67/239.3 *32 = 1.98 g?? but that's wrong?? Help please?

You don't convert anything to grams. Use mols to convert from what you have to what you want. Then convert to grams. Reactions go by mols, not grams.

To calculate the grams of oxygen required to react with 0.124 moles of lead(II) sulfide, you need to use the balanced equation and the stoichiometry of the reaction.

From the balanced equation:
2PbS + O2 -> 2PbO + 2SO2

There is a 1:1 ratio between the moles of lead(II) sulfide (PbS) and moles of oxygen (O2).

First, calculate the molar mass of PbS:
Molar mass of PbS = atomic mass of Pb (207.2 g/mol) + atomic mass of S (32.1 g/mol)
= 239.3 g/mol

Next, calculate the grams of PbS:
Grams of PbS = 0.124 moles x molar mass of PbS
= 0.124 mol x 239.3 g/mol
= 29.67 g

The molar mass of O2 is 32.0 g/mol.

Now, use the stoichiometry of the balanced equation to find the grams of oxygen required.

From the balanced equation, it can be seen that for every 2 moles of PbS, 1 mole of O2 is required.

So, the ratio is:
2 moles of PbS: 1 mole of O2

Therefore, the moles of oxygen required can be calculated as follows:
Moles of O2 = Moles of PbS / Ratio of O2 to PbS
= 0.124 mol PbS / 2 mol PbS:1 mol O2
= 0.062 mol O2

Finally, calculate the grams of oxygen required:
Grams of O2 = Moles of O2 x Molar mass of O2
= 0.062 mol O2 x 32.0 g/mol
= 1.984 g

Therefore, approximately 1.984 grams of oxygen are required to react with 0.124 moles of lead(II) sulfide.

To determine the amount of oxygen required to react with a given amount of lead(II) sulfide, you need to follow a few steps:

1. Convert the given moles of lead(II) sulfide (PbS) to grams using its molar mass. You correctly calculated this step: 0.124 mol PbS * molar mass of PbS (239.3 g/mol) = 29.67 g PbS.

2. Next, use the coefficients from the balanced equation to establish the mole ratio between PbS and O2. From the balanced equation, you can see that the ratio is 2 moles of PbS : 1 mole of O2.

3. Convert the grams of PbS to moles of PbS using its molar mass. 29.67 g PbS * (1 mol PbS / 239.3 g PbS) ≈ 0.124 mol PbS.

4. Use the mole ratio from step 2 to find the moles of O2 required. Since the mole ratio is 2 PbS : 1 O2, you can calculate: 0.124 mol PbS * (1 mol O2 / 2 mol PbS) = 0.062 mol O2.

5. Finally, convert moles of O2 to grams using its molar mass. The molar mass of O2 is 32 g/mol. Therefore, 0.062 mol O2 * (32 g O2 / 1 mol O2) = 1.984 g O2.

So, the correct amount of oxygen required to react with 0.124 moles of lead(II) sulfide is approximately 1.984 grams.