If 127g of copper reacts with 32g of oxygen gas to for copper oxide how much copper oxide is produced?

Answer 1

Copper oxide can exist as Cu2(I)O or CU(II)O. You need to find this out. It looks like Oxygen will limit the reaction anyway. Find the amount of moles by dividing the molecular mass of the limiting reagent, apply stochiometry, then find weight of product

Answer 2

To find the amount of copper oxide produced, we need to determine the limiting reactant. The balanced equation is 4Cu + O2 → 2Cu2O. Calculate the moles of each reactant: 127g Cu / 63.5 g/mol Cu = 2 moles Cu, and 32g O2 / 32 g/mol O2 = 1 mole O2. Since 1 mole of O2 is needed for 4 moles of Cu, O2 is the limiting reactant. Hence, 2 moles Cu will react with 1 mole O2 to produce 2 moles Cu2O, which is 63.5 x 2 x 2 = 254g Cu2O.

Answer 3

If 127g of zinc is reacted with 160g of copper oxide what is the mass of the products formed?