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

C.) K is limiting, 11.2 g of formed

Explanation:

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

Putting in the values we get:

According to stoichiometry :

4 moles of require 1 mole of

Thus 0.238 moles of will require=  of

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

As 4 moles of give = 2 moles of

Thus 0.238  moles of will give =  of

Mass of

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

it should be p1 v1 = p2 v2

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.