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!
A. You have 100/95.6 = 1.047 moles of CuS and 56/32 = 1.75 moles of O2.
Since the number of moles of each reactant consumed in the reaction are the same,there will be excess O2. CuS is the limiting reactant.
B. 1.75-1.046 = ? moles of O2 will remain unreacted. Convert this to grams.
To determine the limiting reactant and the excess reactant in this problem, you need to compare the amount of each reactant to the stoichiometry of the balanced equation.
A. To determine the limiting reactant, we need to calculate the number of moles of each reactant and compare them to the stoichiometric ratio.
1. Calculate the number of 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
2. Calculate the number of moles of O2 using the same method:
molar mass of O2 = 2 * atomic mass of O
= 2 * 16.00 g/mol
= 32.00 g/mol
moles of O2 = 56 g / 32.00 g/mol
3. Compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation (2:2):
moles ratio = moles of CuS / 2 (stoichiometric coefficient of CuS)
= moles of O2 / 2 (stoichiometric coefficient of O2)
If the moles ratio is less than 1, then CuS is the limiting reactant. If the moles ratio is greater than 1, then O2 is the limiting reactant.
B. To determine the excess reactant and the remaining mass:
1. Once you've identified the limiting reactant in part A, calculate the mass of product formed using the stoichiometry of the balanced equation.
Calculate the number of moles of the limiting reactant:
moles of limiting reactant = moles of limiting reactant (from part A)
Use the stoichiometry to calculate the moles of product formed (CuO) from the moles of limiting reactant:
moles of CuO = moles of limiting reactant * (2 moles of CuO / 2 moles of CuS)
Calculate the mass of CuO formed using the molar mass of CuO:
mass of CuO formed = moles of CuO * molar mass of CuO
2. Determine the amount of excess reactant by subtracting the moles of limiting reactant from the moles of excess reactant. Then, calculate the remaining mass of the excess reactant using its molar mass.
Determine the moles of excess reactant:
moles of excess reactant = moles of excess reactant (from part A) - moles of limiting reactant
Calculate the remaining mass of the excess reactant:
mass of excess reactant remaining = moles of excess reactant * molar mass of excess reactant
By following these steps, you will be able to determine the limiting reactant, the excess reactant, and the remaining mass after the reaction is completed.