# 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?

May 14, 2018

Cu is the limiting reactant
$5 g \text{ } {O}_{2}$ in excess
$\approx 15 g \text{ } C u {O}_{2}$ produced

#### Explanation:

Given: $10$ g Cu reacts with $10$ g ${O}_{2}$.

Skeletal equation: $\text{ } C u + {O}_{2} \to C u {O}_{2}$

balanced equation is the same as the skeletal equation.

molar masses:

$63.546 \frac{g}{m o l} \text{ "Cu; " " 2(15.999) = 31.998 g/(mol) " } {O}_{2}$

$63.546 + 31.998 = 95.544 \frac{g}{m o l} \text{ } C u {O}_{2}$

Find the limiting reagent:

$10 g C u \times \frac{1 m o l C u}{63.546 g C u} \times \frac{1 m o l C u {O}_{2}}{1 m o l C u} = .1574 m o l \text{ } C u {O}_{2}$

$10 g {O}_{2} \times \frac{1 m o l {O}_{2}}{31.998 g {O}_{2}} \times \frac{1 m o l C u {O}_{2}}{1 m o l C u} = .1325 m o l \text{ } C u {O}_{2}$

Cu is the limiting reagent.

${O}_{2}$ is in excess.

Find the amount of ${O}_{2}$ used:
$.1574 m o l \text{ " CuO_2 xx (1 mol O_2)/(1 mol CuO_2) xx (31.998 g O_2)/(1 mol O_2) ~~ 5 g " } {O}_{2}$

Find the mass of ${O}_{2}$ in excess:
$10 g - 5 g \left(\text{used}\right) = 5 g$ in excess

Find the mass of produced product $C u {O}_{2}$:

$.1574 m o l \text{ " CuO_2 xx (95.544 g CuO_2)/(1 mol CuO_2) = 15.04 g " } C u {O}_{2}$