Asked by Lucas
How do I solve this: 2Cu2O(solid) + Cu2S(solid) --> 6Cu(solid) + SO2(gas). 86g of copper 1 oxide reacts with 46g of copper 1 sulfide. Find mass of Copper made?
Answers
Answered by
DrBob222
2Cu2O(s) + Cu2S(s) --> 6Cu(s) + SO2(g)
This is a limiting reagent (LR) problem. You know that when more than one reactant has a amount listed. Two ways. I'll do the long way. Easier to explain.
mols Cu2O = g/molar mass = 86.0/molar mass Cu2O = about 0.6 mole but you use a more accurate value.
mols Cu2S = 46.0/molar mass Cu2S = 0.9
Next you determine how many moles you will get of Cu for EACH. You use the coefficient in the balanced equation to convert from moles reactant to mols product.
For Cu2O that is 0.6 mol Cu2O x (6 mol Cu/2 Cu2O) = about 0.6 x 3 = about 1.8
For Cu2S that is 0.9 x mol Cu2S x (6 mol Cu/1 mol Cu2S) = about 0.9 x 6 = about 5.4
For LR reactions the smaller amount always wins; therefore, you will obtain 1.8 moles Cu. Grams Cu x atomic mass Cu = ?
Post your work if you get stuck. Remember to clean up those number above to get your final answer.
This is a limiting reagent (LR) problem. You know that when more than one reactant has a amount listed. Two ways. I'll do the long way. Easier to explain.
mols Cu2O = g/molar mass = 86.0/molar mass Cu2O = about 0.6 mole but you use a more accurate value.
mols Cu2S = 46.0/molar mass Cu2S = 0.9
Next you determine how many moles you will get of Cu for EACH. You use the coefficient in the balanced equation to convert from moles reactant to mols product.
For Cu2O that is 0.6 mol Cu2O x (6 mol Cu/2 Cu2O) = about 0.6 x 3 = about 1.8
For Cu2S that is 0.9 x mol Cu2S x (6 mol Cu/1 mol Cu2S) = about 0.9 x 6 = about 5.4
For LR reactions the smaller amount always wins; therefore, you will obtain 1.8 moles Cu. Grams Cu x atomic mass Cu = ?
Post your work if you get stuck. Remember to clean up those number above to get your final answer.
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