# What mass of CuO can be formed from the reaction of 18.7 g of CuS and 12.0 g of O2? 2CuS +3O2 → 2CuO + 2SO2

I don't know how to go about solving this problem. The assignment is under solving percent yields and limiting reagents? but they didn't teach me that? How do I apply it to this? How do I start?

No. of moles of O2 = 12/ (2 x 16) = 0.375 mol

No. of moles of CuS = 18.7/ (63.5 + 32.1) = 0.196 mol

From balanced equation, 0.196 mol of CuS only requires 0.294 [(0.196/2) x 3] mol of O2. Hence, CuS is the limiting reagent.

No. of CuO = 0.196 mol

Mass of CuO = 0.196 x (63.5 + 16) = 15.55 g

The concept used was to find out which is the limiting reagent and from there using mole ratio to find mass of CuO.
Ok so you have the reactants (2CuS + 3O2) and then the products (2CuO + 2SO2). The question asks how much of the product CuO can be made. Well because we know CuS = 18.7 g and 02 = 12.0 g we can then find out how much CuO can be made. That's all I can tell you, I don't want to give you a wrong answer.

hope that it'll help

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