Convert to moles -- divide 49.1 by the molecular weight of Cu2O (143 g /mole but you may want more significant figures in these number) to find moles of Cu2O. The divide 44.1 g by 32 g/mole to find moles of O2. Remember that it takes two moles of Cu2O for every mole of O2 and determine the limiting reagent. Now compute the number of moles of CuO you will make (again, remembering the stoichiometry of the balanced equation). Now multiply the moles of CuO by the molecular weight of CuO and you have the answer, in grams.

CuO and Cu2O -- Limiting reactant problem.....

The oxides of copper can be produced by the heating of copper metal in the presence in oxygen. A mixture is produced with mild heating. Thorough heating with an excess of oxygen will form copper(II) oxide.

One say to solve limiting reactant problems is to compute the amount of product using each of the reactants. The theoretical yield will be the lesser amount, which also tells you the limiting reactant.

2Cu2O(s) + O2(g) --> 4CuO(s)

49.1g ......... 44.1g ...... ??? g

49.1g Cu2O x (1 mol Cu2O / 143.1g Cu2O) x ( 2 mol CuO / 1 mol Cu2O) x (79.55g CuO / 1 mol CuO) = 54.6g CuO

44.1g O2 x (1 mol O2 / 32.0g O2) x (4 mol CuO / 1 mol O2) x (79.55g CuO / 1 mol CuO) = 438g CuO

The theoretical yield is 54.6g CuO and Cu2O is the limiting reactant.