2Cu2O(s) + O2(g) → 4CuO(s) Determine the mass (in g) of CuO formed if 61.6 g of Cu2O reacts with 56.6 g of O2. g CuO

Answer 1

Haley,

You are asking a lot of the same kind of questions. Hopefully you are understanding how to approach these. We won't be there to help on the test. ALWAYS WORK IN MOLES, then CONVERT TO GRAMS.

2Cu₂O + O₂ ==> 4CuO ... balanced equation

moles of Cu₂O = 61.6 g x 1 mole Cu₂O/143 g = 0.4308 moles Cu₂O

moles O₂ = 56.6 g O₂ x 1 mole/32 g = 1.769 moles O₂

Now you need to determine which reactant is limiting. Easiest way is to divide each by it's coefficient and see which is the least. Thus, 0.4308/2 = 0.2154 for Cu₂O and 1.769/1 = 1.769 for O₂ so Cu₂O IS LIMITING

Since Cu₂O is limiting, it will determine moles of CuO than can be formed.

Moles CuO formed = 0.4308 moles Cu₂O x 4 moles CuO/2 moles Cu₂O = 0.8616 moles CuO

Mass (grams) CuO formed = 0.8616 moles CuO x 79.55 g/mole = 68.5 g (to 3 sig. figs.)