A. You have 100/95.6 = 1.047 moles of CuS and 56/32 = 1.75 moles of O2.
Since the number of moles of each reactant consumed in the reaction are the same,there will be excess O2. CuS is the limiting reactant.
B. 1.75-1.046 = ? moles of O2 will remain unreacted. Convert this to grams.
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!
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