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, you need to compare the number of moles of each reactant to the stoichiometry of the balanced equation.
A.
1. Calculate the number of moles of CuS (copper(II) sulfide):
- Moles = mass / molar mass
- Molar mass of CuS = atomic mass of Cu + atomic mass of S
- Atomic mass of Cu = 63.55 g/mol
- Atomic mass of S = 32.07 g/mol
- Moles of CuS = 100 g / (63.55 g/mol + 32.07 g/mol)
2. Calculate the number of moles of O2 (oxygen gas):
- Moles = mass / molar mass
- Molar mass of O2 = 2 * atomic mass of O
- Atomic mass of O = 15.999 g/mol
- Moles of O2 = 56 g / (2 * 15.999 g/mol)
3. Compare the moles of CuS and O2 to the stoichiometry of the balanced equation:
- From the balanced equation, you can see that the stoichiometric ratio is 2:2 between CuS and O2.
- Divide the moles of CuS by the stoichiometric coefficient of CuS:
Moles of CuS / Stoichiometric coefficient of CuS = Moles of CuS / 2
- Divide the moles of O2 by the stoichiometric coefficient of O2:
Moles of O2 / Stoichiometric coefficient of O2 = Moles of O2 / 2
- Compare the two values obtained:
If Moles of CuS / 2 < Moles of O2 / 2, then CuS is the limiting reactant.
If Moles of CuS / 2 > Moles of O2 / 2, then O2 is the limiting reactant.
B.
1. Determine the limiting reactant (as found in part A).
2. To find out the excess reactant, calculate the number of moles reacted based on the stoichiometry of the balanced equation:
- Based on the limiting reactant, determine the number of moles of that reactant that reacts. For example, if CuS is the limiting reactant:
Moles of CuS reacted = (Moles of CuS / 2) * Stoichiometric coefficient of CuS
Then, calculate the remaining moles of CuS:
Remaining moles of CuS = Moles of CuS - Moles of CuS reacted
- Repeat the same process if O2 is the limiting reactant.
- Convert the remaining moles of the excess reactant to grams:
Remaining grams = Remaining moles * Molar mass of excess reactant.
So, after following these steps, you will determine which reactant is limiting, and in the case of excess reactant, you will find the amount of grams remaining after the reaction is completed.