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 reactant is in excess, we need to compare the amount of each reactant available to the stoichiometric ratio of the balanced equation.
Let's start with reactant A, which is CuS (copper(II) sulfide). The molar mass of CuS is 95.6 g/mol (63.5 g/mol for copper + 32.1 g/mol for sulfur). The molar mass of O2 is 32 g/mol.
A. To find out which reactant is limiting, we need to calculate the number of moles of each reactant using their respective masses.
Moles of CuS = (mass of CuS) / (molar mass of CuS)
= 100 g / 95.6 g/mol
= 1.05 mol (rounded to two decimal places)
Moles of O2 = (mass of O2) / (molar mass of O2)
= 56 g / 32 g/mol
= 1.75 mol (rounded to two decimal places)
Now we can compare the stoichiometric ratio of CuS and O2 in the balanced equation:
From the balanced equation: 2CuS + 2O2 -> 2CuO + 2CO2
The stoichiometric ratio between CuS and O2 is 2:2, which means they react in a 1:1 ratio.
Since we have 1.05 moles of CuS and 1.75 moles of O2, it indicates that we have more moles of O2 than CuS.
Therefore, CuS is the limiting reactant.
B. To determine the excess reactant and how many grams remain after the reaction, we can use the limiting reactant (CuS) to find the theoretical yield of CuO (copper(II) oxide).
From the stoichiometry of the balanced equation, we know that 2 moles of CuS react to form 2 moles of CuO.
Moles of CuO = (moles of CuS) x (2 moles CuO / 2 moles CuS)
= 1.05 mol x (2 mol CuO / 2 mol CuS)
= 1.05 mol
Now, we can calculate the mass of CuO formed using its molar mass, which is 79.5 g/mol (63.5 g/mol for copper + 16 g/mol for oxygen).
Mass of CuO = (moles of CuO) x (molar mass of CuO)
= 1.05 mol x 79.5 g/mol
= 83.48 g (rounded to two decimal places)
Since CuS is the limiting reactant, we can calculate the amount of excess reactant (O2) remaining after the reaction:
Moles of O2 used in the reaction: (moles of CuS) x (2 moles O2 / 2 moles CuS)
= 1.05 mol
Moles of O2 remaining: (moles of O2 initial) - (moles of O2 used)
= 1.75 mol - 1.05 mol
= 0.70 mol (rounded to two decimal places)
To find the mass of O2 remaining, we use its molar mass:
Mass of O2 remaining: (moles of O2 remaining) x (molar mass of O2)
= 0.70 mol x 32 g/mol
= 22.40 g (rounded to two decimal places)
Therefore, the excess reactant (O2) is in excess by 22.40 g, and this amount remains after the reaction is completed.