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 solve these types of problems, we need to compare the given amounts of each reactant to their respective stoichiometric ratios in the balanced equation. Let's go step by step.
A. To determine the limiting reactant, we need to compare the moles of each reactant to their stoichiometric ratios.
1. Calculate the moles of CuS:
- Given mass of CuS = 100 g
- Molar mass of CuS = 63.55 g/mol (Cu) + 32.06 g/mol (S) = 95.61 g/mol
- Moles of CuS = mass / molar mass = 100 g / 95.61 g/mol = 1.046 mol
2. Calculate the moles of O2:
- Given mass of O2 = 56 g
- Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol
- Moles of O2 = mass / molar mass = 56 g / 32.00 g/mol = 1.75 mol
3. Now, compare the moles of CuS and O2 to their stoichiometric ratios in the balanced equation.
Stoichiometric ratio for CuS:O2 = 2:2 = 1:1
Moles of CuS = 1.046 mol
Moles of O2 = 1.75 mol
Since the stoichiometric ratio is the same, we can see that the mole ratio matches, which means both reactants are in equal proportions. Therefore, none of the reactants is limiting in this case.
B. To determine which reactant is in excess and calculate the grams remaining, we need to compare the moles of the reactants used.
1. Calculate the moles of CuO produced:
- According to the balanced equation, 2 moles of CuS produce 2 moles of CuO.
- Moles of CuO = 1.046 mol
2. Calculate the moles of CO2 produced:
- According to the balanced equation, 2 moles of CuS produce 2 moles of CO2.
- Moles of CO2 = 1.046 mol
3. Calculate the amount of CuS and O2 remaining.
CuS remaining = initial moles of CuS - moles of CuO produced
= 1.046 mol - 1.046 mol = 0 mol
O2 remaining = initial moles of O2 - moles of CO2 produced
= 1.75 mol - 1.046 mol = 0.704 mol
4. Calculate the grams of CuS and O2 remaining.
CuS remaining = 0 mol
Molar mass of CuS = 95.61 g/mol (as calculated earlier)
O2 remaining = 0.704 mol
Molar mass of O2 = 32.00 g/mol (as calculated earlier)
Grams of CuS remaining = 0 mol * 95.61 g/mol = 0 g
Grams of O2 remaining = 0.704 mol * 32.00 g/mol = 22.53 g
Therefore, in this reaction, all of the CuS is consumed, and 22.53 grams of O2 remain after the reaction is completed.