Question

How many grams of oxygen gas are contained in a 15 L sample at 1.02 atm and 28°C? Please show your work.

Answer 1

Answer:

0.019 g.

Explanation:

• Firstly, we need to find the no. of moles of oxygen gas:

• We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.02 atm).

V is the volume of the gas in L (V = 15.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 28°C + 273 = 301.0 K).

∴ n = PV/RT = (1.02 atm)(15.0 L)/(0.0821 L.atm/mol.K)(301.0 K) = 0.62 mol.

• To find the mass of oxygen gas, we have:

no. of moles = mass/molar mass.

∴ mass of oxygen = (no. of moles)(molar mass) = (0.62 mol)(32.0 g/mol) = 0.019 g.