Question

PLEASE HELP!!!!
If 9.30 g of potassium reacts with 2.50 g of O2 to form K2O , what is the limiting reagent and what is the theoretical yield of the reaction?

Hint: write the balanced reaction

K - 39.10 g/mol

O - 15.999 g/mol

ANSWER CHOICES:

A.) O2 is limiting, 11.2 g of K2O formed


B.) K is limiting, 14.7 g of K2O formed


C.) K is limiting, 11.2 g of K2O formed


D.) O2 is limiting, 14.7 g of K2O formed


E.) O2 is limiting, 19.2 g of K2O formed

Answer 1

Answer: C.) K is limiting, 11.2 g of $K_2O$ formed

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Putting in the values we get:

$\text{Number of moles of potassium}=\frac{9.30g}{39.10g/mol}=0.238moles$

$\text{Number of moles of oxygen}=\frac{2.50g}{31.99g/mol}=0.0781moles$

$4K+O_2\rightarrow 2K_2O$

According to stoichiometry :

4 moles of $K$ require 1 mole of $O_2$

Thus 0.238 moles of $K$ will require=$\frac{1}{4}\times 0.238=0.0595moles$  of $O_2$

Thus $K$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

As 4 moles of $K$ give = 2 moles of $K_2O$

Thus 0.238  moles of $K$ will give =$\frac{2}{4}\times 0.238=0.119moles$  of $AgCl$

Mass of $K_2O=moles\times {\text {Molar mass}}=0.119moles\times 94.2g/mol=11.2g$

Thus K is limiting and 11.2 g of $K_2O$ will be formed.