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 the limiting reactant and the excess reactant, you need to compare the ratio of moles of each reactant to the stoichiometric ratio in the balanced equation.
A. To find the limiting reactant, you need to convert the masses of CuS and O2 into moles.
1. Calculate the moles of CuS:
- Molar mass of CuS (Copper(II) sulfide) = atomic mass of Cu + atomic mass of S
- Molar mass of CuS = (63.55 g/mol) + (32.06 g/mol) = 95.61 g/mol
- Moles of CuS = mass of CuS / molar mass of CuS
- Moles of CuS = 100 g / 95.61 g/mol
2. Calculate the moles of O2:
- Molar mass of O2 (Oxygen) = 2 * atomic mass of O
- Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol
- Moles of O2 = mass of O2 / molar mass of O2
- Moles of O2 = 56 g / 32.00 g/mol
B. Determine the limiting reactant:
1. Use the stoichiometric ratio from the balanced equation (2:2) to compare moles of CuS and O2.
- Dividing the moles of CuS by its stoichiometric coefficient gives the moles required for O2 to react.
- Moles of CuS required to react with O2 = (Moles of O2) / 2
2. Compare the calculated moles of CuS required to react with O2 with the actual moles of CuS available.
- If the calculated moles of CuS required is greater than the actual moles available, then CuS is the limiting reactant.
- If the calculated moles of CuS required is less than or equal to the actual moles available, then O2 is the limiting reactant.
C. Determine the excess reactant and the grams remaining after the reaction:
1. Once you have identified the limiting reactant, you can use the stoichiometry to calculate the moles of the product formed (CuO).
- Moles of CuO = (Moles of limiting reactant) * (Coefficient ratio of CuO from balanced equation)
2. Convert the moles of CuO to grams using its molar mass:
- Molar mass of CuO (Copper(II) oxide) = atomic mass of Cu + atomic mass of O
- Molar mass of CuO = (63.55 g/mol) + (16.00 g/mol) = 79.55 g/mol
- Mass of CuO formed = Moles of CuO * Molar mass of CuO
3. Calculate the excess reactant by subtracting the moles of the limiting reactant consumed from the moles of the excess reactant available.
4. Finally, calculate the grams remaining by converting the moles of the excess reactant to grams using its molar mass.
By following these steps, you should be able to determine the limiting reactant and the excess reactant, as well as calculate the grams remaining after the reaction is completed.