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, you need to compare the amount of each reactant you have to the stoichiometric ratio in the balanced equation.
A. To find the limiting reactant, you need to calculate the number of moles of each reactant.
1. Calculate the number of moles of CuS:
moles = mass / molar mass
molar mass of CuS = (63.5 g/mol) + (32 g/mol) = 95.5 g/mol
moles of CuS = 100 g / 95.5 g/mol = 1.05 mol (rounded to two decimal places)
2. Calculate the number of moles of O2:
moles = mass / molar mass
molar mass of O2 = 32 g/mol
moles of O2 = 56 g / 32 g/mol = 1.75 mol (rounded to two decimal places)
Now, let's compare the ratio of CuS to O2 in the balanced equation.
From the equation, you can see that the ratio is 2:2, which simplifies to 1:1.
So, for every 1 mol of CuS, you need 1 mol of O2.
Looking at the moles we calculated, we have 1.05 mol of CuS and 1.75 mol of O2. Since the ratio is 1:1, O2 is in excess, and CuS is the limiting reactant.
B. To find the reactant in excess and the amount remaining, you need to calculate the amount of excess reactant used and the amount remaining after the reaction is completed.
1. Calculate the moles of CuO produced:
From the balanced equation, you can see that for every 2 moles of CuS, you produce 2 moles of CuO.
Since CuS is the limiting reactant, we can use its moles (1.05 mol) to calculate the moles of CuO produced.
The ratio of CuS to CuO in the equation is 2:2, so 1.05 mol of CuS will produce 1.05 mol of CuO.
2. Calculate the moles of O2 used:
Again, since CuS is the limiting reactant, we'll use its moles (1.05 mol) to calculate the moles of O2 used.
The ratio of CuS to O2 in the equation is 2:2, so 1.05 mol of CuS will require 1.05 mol of O2.
3. Calculate the amount of O2 remaining:
Initially, we had 1.75 mol of O2, and we used 1.05 mol for the reaction.
The remaining moles of O2 = initial moles of O2 - moles of O2 used = 1.75 mol - 1.05 mol = 0.70 mol.
Finally, calculate the mass of O2 remaining:
mass = moles x molar mass
molar mass of O2 = 32 g/mol
mass of O2 remaining = 0.70 mol x 32 g/mol = 22.4 g (rounded to two decimal places)
Therefore:
- The limiting reactant is CuS.
- The reactant in excess is O2.
- After the reaction is completed, 22.4 grams of O2 will remain.