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

Step-by-step 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}=(9.30g)/(39.10g/mol)=0.238moles`

`\text{Number of moles of oxygen}=(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=
`(1)/(4)* 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 =
`(2)/(4)* 0.238=0.119moles` of
`AgCl`

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

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