Question

A sample of propane gas (C3H8) having volume of 1.80L at 25oC and 1.65atm was mixed with a sample of oxygen gas having a volume of 35.0L at 31oC and 1.25atm. The mixture was ignited to form carbon dioxide and water. Write the balanced reaction. What is the limiting reagent? What is the maximum number of grams of water which can form from this reaction? Find the number of moles of excess reactant at STP.

Answer 1

Balance chemical equation can be writen as follows,

C3H8 + 5O2 ==> 3CO2 + 4H2O

Volume of propane 1.8 L at 25^oC and 1.65 atm

Volume of Oxygen 35.0L at 31^oC and 1.25 atm

Now we know that PV=nRT

R= 0.08205 L atm K⁻¹ mol⁻¹

n = PV/RT = 1.65 × 1.8/ 0.08205 × 298.15 = 0.1214 moles of Propane

Similarly calculate number of moles for oxygen, n= 35 × 1.25/ 0.08205 × 304.15 = 1.75 moles of Oxygen

now we know that 1 mole of propane gas needs 5 moles of oxygen to produce 3 moles of CO2 and 4 moles of water.

The reactant that is fully consumed during the reaction is called the limiting reagent. In combustion reactions, oxygen gas is usually present in excess in order for the the reaction to under complete combustion. Complete combustion means that the products of the reaction are CO₂ (g) and H₂O (l). In this given combustion reaction 2 moles of oxygen stays in excess so here limiting reagent is propane.

from the reaction we know that 4 moles of water produce from 1 mole of propane

So,= 0.1214 × 4 × 18.g/mol = 8.7 g of water produce

The limiting reagent is the reactant that is completely used up during the chemical reaction. The reactant that is in excess is the reactant that is not completely used up during the chemical reaction, that is, there is some of this reactant left over.

From the balance equation mole ratio of the reactant is ,C3H8:O2:: 1:5

If all 0.1214 moles of propane to be used in the reaction it would require 5 x 0.1214= 0.607 moles of oxygen for the reaction to go to completion.But there are 1.75 moles of oxygen present which is more that 0.607 moles.

If all of the1.75 moles of oxygen were to be used in the reaction it would require 1/5 x 1.75= 0.35 moles of propane.but there are 0.1214 moles are present which is less that 0.35 moles.

So the reactant is in excess is oxygen: 1.75-0.607 = 1.4 moles of oxygen is in excess.

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