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 reactant is in excess, we need to compare the amount of the reactants with the stoichiometric ratios in the balanced equation.
A. To find the limiting reactant, we can calculate the number of moles of each reactant using their molar masses and then compare their ratios in the balanced equation.
1. Calculate the number of moles of CuS:
Moles of CuS = mass / molar mass
Moles of CuS = 100 g / 95.61 g/mol (molar mass of CuS)
Moles of CuS ≈ 1.047 moles
2. Calculate the number of moles of O2:
Moles of O2 = mass / molar mass
Moles of O2 = 56 g / 32 g/mol (molar mass of O2)
Moles of O2 = 1.75 moles
3. Compare the moles of CuS and O2 using the coefficients from the balanced equation.
From the balanced equation: 2CuS + 2O2 → 2CuO + 2CO2
The ratio of CuS to O2 is 2:2, which means 1 mole of CuS reacts with 1 mole of O2.
Moles of CuS : Moles of O2 = 1.047 : 1.75
Since the ratio is not 1:1, we can determine that O2 is the limiting reactant because it is present in a smaller quantity.
B. To find the reactant in excess and the remaining mass after the reaction, we need to calculate the theoretical amount of the product formed based on the limiting reactant.
1. Calculate the number of moles of CuO formed:
From the balanced equation: 2CuS + 2O2 → 2CuO + 2CO2
Since O2 is the limiting reactant, the number of moles of CuO formed will be equal to the number of moles of O2 used.
Moles of CuO = Moles of O2 = 1.75 moles
2. Calculate the mass of CuO formed:
Mass of CuO = Moles of CuO × molar mass of CuO
Mass of CuO = 1.75 moles × 79.55 g/mol (molar mass of CuO)
Mass of CuO ≈ 139.36 g
3. Determine the excess reactant and calculate the remaining mass:
The excess reactant is CuS since O2 is limiting. To find the remaining mass of CuS, we need to subtract the amount used (based on the stoichiometry) from the initial amount.
Moles of CuS used = Moles of O2 used (since the stoichiometry is 1:1)
Moles of CuS remaining = Moles of CuS initial - Moles of CuS used
Moles of CuS remaining = 1.047 moles - 1.047 moles (assuming complete reaction)
Moles of CuS remaining = 0 moles
Mass of CuS remaining = Moles of CuS remaining × molar mass of CuS
Mass of CuS remaining = 0 moles × 95.61 g/mol (molar mass of CuS)
Mass of CuS remaining = 0 g
Therefore, after the reaction is completed, no CuS remains.
To summarize:
A. O2 is the limiting reactant.
B. CuS is in excess, and 0 grams of CuS remain after the reaction is completed.