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 and which is in excess, we need to compare the amounts of reactants available to the stoichiometric ratio in the balanced equation.
A. To determine the limiting reactant, we need to calculate the moles of each reactant.
1. Calculate the moles of CuS:
Moles of CuS = (mass of CuS)/(molar mass of CuS)
Molar mass of CuS = atomic mass of Cu + atomic mass of S
= (63.55 g/mol) + (32.07 g/mol)
= 95.62 g/mol
Moles of CuS = (100 g)/(95.62 g/mol) ≈ 1.05 mol
2. Calculate the moles of O2:
Moles of O2 = (mass of O2)/(molar mass of O2)
Molar mass of O2 = 32.00 g/mol
Moles of O2 = (56 g)/(32.00 g/mol) = 1.75 mol
3. Use the coefficients from the balanced equation to find the ratio of moles of CuS to O2:
CuS : O2 = 2 : 2 = 1 : 1
Based on the stoichiometric ratio, we can see that both reactants have a 1:1 ratio. Therefore, the limiting reactant is the one with the smaller number of moles, which in this case is CuS.
B. To determine the excess reactant, we need to calculate the moles of the limiting reactant and use it to calculate the moles of the excess reactant.
1. Moles of CuO produced:
From the balanced equation, we can see that 2 moles of CuS react to produce 2 moles of CuO.
Since 1.05 moles of CuS are available (the amount of the limiting reactant), we expect the same number of moles of CuO to be formed.
2. Moles of O2 required to react with all of the CuS:
From the balanced equation, 2 moles of CuS react with 2 moles of O2.
If 1.05 moles of CuS reacted, we would need the same number (1.05 moles) of O2.
3. Calculate the excess moles of O2:
Excess moles of O2 = (moles of O2 available) - (moles of O2 required)
= 1.75 mol - 1.05 mol
= 0.70 mol
4. Calculate the mass of excess O2:
Mass of excess O2 = (excess moles of O2) × (molar mass of O2)
= 0.70 mol × (32.00 g/mol)
≈ 22.4 g
Therefore, the excess reactant is O2, and after the reaction is completed, approximately 22.4 grams of O2 would remain.