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 amount of product that can be formed from each reactant.

A. To find the limiting reactant, we need to calculate the number of moles of each reactant.

First, calculate the molar mass of CuS:
CuS = (63.55 g/mol) + (32.06 g/mol) = 95.61 g/mol

Next, divide the given mass by the molar mass to find the number of moles of CuS:
Number of moles of CuS = 100 g / 95.61 g/mol ≈ 1.05 mol

The molar mass of O2 is:
O2 = (16.00 g/mol) + (16.00 g/mol) = 32.00 g/mol

Now, divide the given mass by the molar mass to find the number of moles of O2:
Number of moles of O2 = 56 g / 32.00 g/mol ≈ 1.75 mol

Next, divide the number of moles of each reactant by the stoichiometric coefficient in the balanced equation to find the ratio of moles involved in the reaction:
CuS: O2 = 1.05 mol / 2 = 0.525
O2: O2 = 1.75 mol / 2 = 0.875

The ratio tells us that for every 1 mole of CuS, we need 0.525 moles of O2. However, we have more moles of O2 available (0.875) compared to the moles of CuS available (0.525). Therefore, CuS is the limiting reactant.

B. To find the excess reactant, we subtract the number of moles of the limiting reactant that participate in the reaction from the total number of moles initially available.

Number of moles of excess O2 = Moles of O2 initially available - Moles of O2 involved in the reaction

Moles of O2 involved in the reaction = (Moles of CuS involved in the reaction) × (Moles of O2 per mole of CuS)
Moles of CuS involved in the reaction = 1.05 mol
Moles of O2 per mole of CuS = 2 mol O2 / 2 mol CuS = 1 mol O2 / 1 mol CuS

Moles of O2 involved in the reaction = 1.05 mol × (1 mol O2 / 1 mol CuS) = 1.05 mol

Moles of O2 initially available = 1.75 mol

Number of moles of excess O2 = 1.75 mol - 1.05 mol = 0.70 mol

To find the remaining mass of excess O2, multiply the moles of excess O2 by its molar mass:
Mass of excess O2 = Number of moles of excess O2 × Molar mass of O2

Molar mass of O2 = 32.00 g/mol

Mass of excess O2 = 0.70 mol × 32.00 g/mol = 22.4 g

Therefore, the excess reactant is O2, and there are 22.4 grams remaining after the reaction is completed.