Question

Stoichiometry involving gases:

1. How many moles of oxygen gas are present in 4.48L of the gas at 750mm-Hg and 87 degrees C?

2. Propane is used in gas grills and combusted with oxygen to produce carbon dioxide and water vapor. If 2L of propane is combusted, what volume of oxygen is required and what volumes of water vapor and carbon dioxide are produced?

Answer 1

Answer – 1) We are given, V = 4.48 L , T = 87+273 = 360 K ,P = 750/760 = 0.989 atm

We know Ideal gas law

n = PV/RT

= 0.989 atm * 4.48 L/ 0.0821 L.atm.mol^-1.K^-1*360 K

   = 0.150 moles

0.150 moles of oxygen gas are present in 4.48L of the gas at 750mm-Hg and 87 degrees C?

2) We are given, volume = 2.0 L we assume, P = 1.00 atm, and T = 25 +273 = 298 K

Reaction –

C₃H₈ + 5 O₂ ------> 3 CO₂ + 4 H₂O

Now we need to calculate moles of propane

Using the Ideal gas law

n = PV/RT

= 1.0 atm * 2.0 L/ 0.0821 L.atm.mol^-1.K^-1*298 K

   = 0.0817 moles

Now moles of O₂

Form the balanced equation

1 moles of C₃H₈ = 5 mole of O₂

So, 0.0817 moles of C₃H₈ = ?

= 0.409 moles

Moles of CO₂

Form the balanced equation

1 moles of C₃H₈ = 3 mole of CO₂

So, 0.0817 moles of C₃H₈ = ?

= 0.245 moles of CO₂

Moles of H₂O

Form the balanced equation

1 moles of C₃H₈ = 4 mole of H₂O

So, 0.0817 moles of C₃H₈ = ?

= 0.327 moles of H₂O

The volume of O₂

V = 0.409 moles* 0.0821 L.atm.mol^-1.K^-1 * 308 K / 1.00 atm

     = 10.0 L

The volume of CO₂

V = 0.245 moles* 0.0821 L.atm.mol^-1.K^-1 * 308 K / 1.00 atm

     = 6.0 L

The volume of H₂O

V = 0.327 moles* 0.0821 L.atm.mol^-1.K^-1 * 308 K / 1.00 atm

     = 8.0 L