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!
2CuS + 2O2 --> 2CuO + 2CO2
No, not balanced, you have no S on the right and you have 4 O atoms left and 6 O atoms on the right
I believe your original equation is a typo even before the balancing.
To solve this problem, we need to determine which reactant is limiting and which reactant is in excess.
A. To find the limiting reactant, we can use the given amounts of both reactants and convert them to the moles using their respective molar masses.
1. Convert the mass of CuS to moles:
- Molar mass of CuS = 63.55 g/mol (atomic mass of Cu) + 32.07 g/mol (atomic mass of S) = 95.62 g/mol
- Moles of CuS = 100 g / 95.62 g/mol = 1.047 mol (rounding to 3 decimal places)
2. Convert the mass of O2 to moles:
- Molar mass of O2 = 16.00 g/mol (atomic mass of O) × 2 = 32.00 g/mol
- Moles of O2 = 56 g / 32.00 g/mol = 1.750 mol (rounding to 3 decimal places)
3. Use the balanced equation coefficients to compare the moles of the reactants:
- Balanced equation: 2CuS + 2O2 → 2CuO + 2CO2
- The ratio of CuS to O2 in the balanced equation is 2:2, or 1:1
- Therefore, the reactant with the lesser number of moles is the limiting reactant.
In this case, both CuS and O2 have the same number of moles (1.047 mol and 1.750 mol), so they are in a 1:1 ratio. Therefore, neither reactant is limiting in this step.
B. Since neither reactant is limiting, we need to determine which reactant is in excess by comparing the moles used in the equation.
1. We know that 2 moles of CuS react with 2 moles of O2 to produce 2 moles of CuO.
- This means that for every 1 mole of CuS, we need 1 mole of O2.
2. To find the excess reactant, we can compare the moles used and see which reactant has leftover moles.
- O2: Since both reactants have the same number of moles (1.047 mol and 1.750 mol), none of the O2 is left over, as it is used up completely in the reaction.
- CuS: Since there is more CuS present (1.047 mol) than necessary (0.523 mol based on the 1:1 ratio with O2), there are still some moles of CuS remaining after the reaction is completed.
To find the amount of CuS remaining in grams, we can use the molar mass of CuS:
- Molar mass of CuS = 95.62 g/mol
- Mass remaining = Moles remaining × Molar mass
- Mass remaining = (1.047 mol - 0.523 mol) × 95.62 g/mol
Evaluating the expression:
Mass remaining = 0.524 mol × 95.62 g/mol ≈ 50 g (rounded to the nearest gram)
Therefore, the reactant in excess is O2, and 50 grams of CuS will remain after the reaction is completed.