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 in this problem, we need to calculate the number of moles of each reactant and compare them to the stoichiometric ratio in the balanced equation.
A. To determine which reactant is limiting, we need to compare the moles of CuS and O2.
1. Calculate the number of moles of CuS:
Moles of CuS = Mass of CuS (g) / molar mass of CuS
The molar mass of CuS can be calculated using the atomic masses:
Molar mass of CuS = (atomic mass of Cu) + (2 x atomic mass of S)
The atomic mass of Cu is 63.55 g/mol, and the atomic mass of S is 32.07 g/mol.
Therefore, the molar mass of CuS = 63.55 g/mol + (2 x 32.07 g/mol)
2. Calculate the number of moles of O2:
Moles of O2 = Mass of O2 (g) / molar mass of O2
The molar mass of O2 can be calculated using the atomic masses:
Molar mass of O2 = (2 x atomic mass of O)
The atomic mass of O is 16 g/mol.
Therefore, the molar mass of O2 = 2 x 16 g/mol
3. Compare the moles of CuS and O2:
- Divide the moles of CuS by the stoichiometric coefficient in the balanced equation (2 moles).
- Divide the moles of O2 by the stoichiometric coefficient in the balanced equation (2 moles).
4. The reactant with the smaller calculated value is the limiting reactant.
B. To calculate the mass of the excess reactant remaining after the reaction is completed:
1. Determine the moles of the limiting reactant consumed using the stoichiometric ratio from the balanced equation.
2. Calculate the mass of the limiting reactant consumed using the moles of the limiting reactant and its molar mass.
3. Subtract the mass of the limiting reactant consumed from the initial mass of the excess reactant to find the mass remaining.
Let's calculate both parts of the question step by step:
A. Determine the limiting reactant:
1. Moles of CuS = 100 g / molar mass of CuS
(Calculate the molar mass of CuS using the given atomic masses)
2. Moles of O2 = 56 g / molar mass of O2
(Calculate the molar mass of O2 using the given atomic mass)
3. Compare the moles of CuS and O2 by dividing them by their respective stoichiometric coefficients.
(Divide the calculated moles by 2)
4. The reactant with the smaller calculated value is the limiting reactant.
B. Calculate the mass of the excess reactant remaining:
1. Determine the moles of the limiting reactant consumed using the stoichiometric ratio from the balanced equation.
2. Calculate the mass of the limiting reactant consumed using the moles of the limiting reactant and its molar mass.
3. Subtract the mass of the limiting reactant consumed from the initial mass of the excess reactant to find the mass remaining.
By following these steps, you will be able to find the answers to both parts of the problem.