Question

10 g of Cu, 10 g of O2 react. Which is the limiting reactant? What amount of reactant remains in excess? Calculate the number of particles of reactant in excess. What is the mass of the produced product?

Answer 1

Cu is the limiting reactant
`5 g " "O_2` in excess
`~~15 g " "CuO_2` produced

Explanation:

Given: `10` g Cu reacts with `10` g `O_2`.

Skeletal equation: `" "Cu + O_2 -> CuO_2`

balanced equation is the same as the skeletal equation.

molar masses:

`63.546 g/(mol) " "Cu; " " 2(15.999) = 31.998 g/(mol) " "O_2`

`63.546 + 31.998 = 95.544 g/(mol) " "CuO_2`

Find the limiting reagent:

`10 g Cu xx (1 mol Cu)/(63.546 g Cu) xx (1 mol CuO_2)/(1 mol Cu) = .1574 mol " " CuO_2`

`10 g O_2 xx (1 mol O_2)/(31.998g O_2) xx (1 mol CuO_2)/(1 mol Cu) = .1325 mol " " CuO_2`

Cu is the limiting reagent.

`O_2` is in excess.

Find the amount of `O_2` used:
`.1574 mol " " CuO_2 xx (1 mol O_2)/(1 mol CuO_2) xx (31.998 g O_2)/(1 mol O_2) ~~ 5 g " "O_2`

Find the mass of `O_2` in excess:
`10 g - 5 g ("used")= 5 g` in excess

Find the mass of produced product `CuO_2`:

`.1574 mol " " CuO_2 xx (95.544 g CuO_2)/(1 mol CuO_2) = 15.04 g " "CuO_2`