How many grams of oxygen gas are present in 44.8 L of oxygen at STP?

Question

a. 8.0 g

b. 64.0 g

c. 32.0 g

d. 48.0 g

Answer 1

im gonna guess c but idk

Answer 2

b.  64.0 g is the right answer.

this is a straight stoichiometry problem...

44.8 L O2 * (1 mol O2 / 22.4 L O2) * (32.0 g O2 / 1 mol O2) = 64.0 g O2

Answer 3

K_bounds has the right answer, but she is assuming you know one mole of any gas at STP is 22.4 liters.

use the ideal gas equation solving for n (moles), then multiply by the molecular weight of oxygen. Remember oxygen gas is O2 and weighs 32g/mole not 16!

Answer 4

The ideal gas equation can be used to solve this problem. The condition is in STP which means that the temperature is 0deg Celcius or 273.15K and pressure is 1atmosphere.

molar mass of oxygen gas = 32.00g/mol

R = 0.0821 L-atm / mol-K

PV = nRT

n = mass / molar mass

Therefore, PV = (mass / molar mass)RT

mass = (PV x molar mass) / RT

mass = (1 atm x 44.8 L x 32 g/mol) / (0.0821 L-atm / mol-K x 273.15 K)

mass = 63.93 g (approximately equal to 64 g)

The answer is:

B. 64.0 grams