This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
2CuS + 3O2 ==> 2CuO + 2SO2
mols CuS = grams/molar mass
mols O2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols CuS to mols CuO.
Do the same for mols O2 to mols CuO.
It is likely these to values for CuO will not agree which means one is not right; the correct answer in LR problem is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value, convert mols CuO to grams. g = mols x molar mass. This is the theoretical yield(TY). The actual yield (AY) is given in the problem as 33.5 g.
% yield = (AY/TY)*100 = ?
Copper(II) sulfide reacts with oxygen to produce copper (II) oxide plus sulfur dioxide.
Suppose you start this reaction with 50.0 grams of copper(II) sulfide and 50.0 grams of oxygen and you actually produce 33.5 grams of copper(II) oxide.
Calculate the percent yield for the reaction.
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