im a little confuse on this problem.

"in the production of copper from ore containing copper(II) sulfide, the ore is first roasted to change it to the oxide according to the following equation: 2CuS + 2O2 --> 2CuO + 2CO2 [i already balanced it]

A. If 100 g of CuS and 56 g of O2 are available, which reactant is limiting?

B. Which reactant is in excess, and how many grams remain after the reaction is completed?

if you could tell me how you do it would be realllly helpful as well.
Thank you!

To determine which reactant is limiting, we need to compare the number of moles of each reactant with their stoichiometric coefficients in the balanced equation.

First, we need to calculate the number of moles of each reactant using their respective molar masses.

Molar mass of CuS: Cu (63.55 g/mol) + S (32.07 g/mol) = 95.62 g/mol
Number of moles of CuS = Mass of CuS / Molar mass of CuS = 100 g / 95.62 g/mol

Molar mass of O2: O (16.00 g/mol) x 2 = 32.00 g/mol
Number of moles of O2 = Mass of O2 / Molar mass of O2 = 56 g / 32.00 g/mol

Next, we compare the moles of CuS and O2 to their stoichiometric coefficients in the balanced equation.

From the balanced equation: 2CuS + 2O2 --> 2CuO + 2CO2
The stoichiometric coefficient ratio of CuS to O2 is 2:2.

Comparing the number of moles:
CuS:O2 = (100 g / 95.62 g/mol) : (56 g / 32.00 g/mol)

Now, we calculate the ratio:
CuS:O2 = (100 g / 95.62 g/mol) x (32.00 g/mol / 56 g)

Simplifying the ratio:
CuS:O2 = 2.12

Since the ratio is exactly 2, this means that the moles of CuS and O2 are in a 1:1 ratio as required by the balanced equation. Therefore, neither reactant is in excess and both will be fully consumed.

To determine the mass of each reactant that remains after the reaction is completed, you can use stoichiometry. Start with the limiting reactant, CuS, and calculate the mass of the excess reactant, O2, that reacts completely.

We know that 100 g of CuS is used and the balanced equation tells us that 2 moles of CuS react with 2 moles of O2. Therefore, the number of moles of O2 required to react completely with 100 g of CuS is:
(Moles of CuS) x (Moles of O2 / Moles of CuS) = (100 g / 95.62 g/mol) x (2 mol O2 / 2 mol CuS)

Now we can use the number of moles of O2 to calculate the mass of O2 required:
Mass of O2 = Moles of O2 x Molar mass of O2 = (100 g / 95.62 g/mol) x (2 mol O2 / 2 mol CuS) x (32.00 g/mol)

Finally, subtract the mass of O2 that reacts from the initial mass of O2 to determine the mass of O2 remaining.

Once you have these calculations, you can determine which reactant is in excess and how many grams remain after the reaction is completed.