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, you need to calculate the amount of product that can be formed from each reactant. The reactant that produces less product is the limiting reactant, and the reactant that is not fully consumed is in excess.
Let's start with calculating the moles of each reactant:
1) CuS (copper(II) sulfide):
Given mass of CuS = 100 g
Molar mass of CuS = atomic mass of Cu + 2 * atomic mass of S = 63.55 g/mol + 2 * 32.07 g/mol = 95.19 g/mol
Number of moles of CuS = mass of CuS / molar mass of CuS
= 100 g / 95.19 g/mol
≈ 1.05 mol
2) O2 (oxygen):
Given mass of O2 = 56 g
Molar mass of O2 = 2 * atomic mass of O = 2 * 16.00 g/mol = 32.00 g/mol
Number of moles of O2 = mass of O2 / molar mass of O2
= 56 g / 32.00 g/mol
= 1.75 mol
Now, based on the balanced chemical equation, the ratio of moles of CuS to CuO is 2:2, which indicates that 2 moles of CuS react with 2 moles of O2 to produce 2 moles of CuO.
Using this information, let's compare the number of moles of CuS and O2:
The stoichiometric ratio of CuS to CuO is 2:2, which means for every 2 moles of CuS, we would expect to form 2 moles of CuO.
In this case, as 1.05 mol of CuS is less than the expected 2 moles, CuS is the limiting reactant.
Now, to determine the excess reactant (O2), we will use the ratio between the limiting reactant (CuS) and the excess reactant (O2) in the balanced chemical equation:
The stoichiometric ratio of CuS to O2 is 2:2, which means for every 2 moles of CuS, we would expect to need 2 moles of O2.
However, we have only 1.05 mol of CuS (limiting reactant). Therefore, the amount of O2 required for the reaction would be:
(1.05 mol CuS)/(2 mol CuS) x (2 mol O2)/(2 mol CuS) = 1.05 mol O2
This shows that only 1.05 mol of O2 is required for the reaction, but we have 1.75 mol of O2 available. Hence, O2 is the excess reactant.
To determine the remaining excess reactant (O2) after the reaction is completed, we will calculate the number of moles of O2 consumed:
Number of moles of O2 consumed = 1.05 mol (since CuS is the limiting reactant)
And to determine the remaining moles of O2:
Remaining moles of O2 = Initial moles of O2 - Moles of O2 consumed
= 1.75 mol - 1.05 mol
= 0.70 mol
Finally, to find the mass of the remaining O2, we'll use the molar mass of O2:
Mass of remaining O2 = Remaining moles of O2 x molar mass of O2
= 0.70 mol x 32.00 g/mol
= 22.4 g
Therefore, the excess reactant (O2) remains as 22.4 grams after the reaction is completed.