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 this problem, we first need to understand the concept of limiting reactants and excess reactants.
The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed. The reactant that is not completely consumed is called the excess reactant.
Let's start by calculating the number of moles for each reactant:
A. For CuS:
Given mass of CuS = 100 g
Molar mass of CuS = 159.16 g/mol
Number of moles of CuS = mass/molar mass = 100 g/159.16 g/mol
B. For O2:
Given mass of O2 = 56 g
Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = mass/molar mass = 56 g/32.00 g/mol
Now, let's determine the limiting reactant:
To compare the ratios between the reactants, we need to use the balanced equation. From the balanced equation:
2 moles of CuS react with 2 moles of O2
We can calculate the mole ratio between CuS and O2:
Number of moles of CuS/2 = Number of moles of O2/2
Now, substitute the calculated values:
Number of moles of CuS/2 = 100 g/159.16 g/mol /2
Number of moles of O2/2 = 56 g/32.00 g/mol /2
Simplify the equation:
Number of moles of CuS = 0.3141 mol
Number of moles of O2 = 1.75 mol
Since the mole ratio is 1:1, we can see that the limiting reactant is CuS because it has fewer moles compared to O2.
Now, let's determine the excess reactant and the amount of it that remains after the reaction:
To calculate the excess reactant, we need to see how much of the excess reactant is left over after the reaction consumes the limiting reactant.
Using the balanced equation, we know that 2 moles of CuS react with 2 moles of O2. Therefore, 1 mole of CuS will react with 1 mole of O2.
Now, let's calculate the maximum moles of CuS that can react:
Maximum moles of CuS = Number of moles of CuS x (1 mole of O2 / 1 mole of CuS) = 0.3141 mol
Next, we calculate the excess moles of O2:
Excess moles of O2 = Number of moles of O2 - Maximum moles of CuS
Finally, calculate the excess mass of O2:
Excess mass of O2 = Excess moles of O2 x Molar mass of O2
Excess mass of O2 = (Number of moles of O2 - Maximum moles of CuS) x Molar mass of O2
Excess mass of O2 = (1.75 mol - 0.3141 mol) x 32.00 g/mol
Now, you can calculate the excess mass of O2 and the remaining mass after the reaction is completed.
I hope this explanation helps you understand how to solve this problem!