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 solve this problem, we need to determine which reactant is limiting and which one is in excess. We can do this by comparing the stoichiometry of the balanced equation with the given amounts of reactants.
A. To determine the limiting reactant, we can use the concept of stoichiometry. The balanced equation tells us that 2 moles of CuS react with 2 moles of O2 to form 2 moles of CuO and 2 moles of CO2. Therefore, the ratio of moles of CuS to moles of O2 is 1:1.
Given that 100 g of CuS and 56 g of O2 are available, we can find the moles of each reactant by using their molar masses. The molar mass of CuS is 159.16 g/mol, so the number of moles of CuS is:
moles of CuS = 100 g / 159.16 g/mol ≈ 0.6288 mol
The molar mass of O2 is 32.00 g/mol, so the number of moles of O2 is:
moles of O2 = 56 g / 32.00 g/mol = 1.75 mol
Since the ratio of moles of CuS to moles of O2 is 1:1, it means that CuS and O2 have the same number of moles available. Therefore, neither CuS nor O2 is limiting. They are present in equal amounts, and both will be completely consumed in the reaction.
B. Since both reactants are in equal amounts, neither reactant will be in excess. Thus, there will be no reactant remaining after the reaction is completed.
To summarize:
A. Neither reactant is limiting.
B. Neither reactant is in excess, and no reactant will remain after the reaction is completed.
I hope this explanation helps! Let me know if you have any further questions.