Question

The burning of propane gas can be represented as a balanced chemical reaction as follows: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) Calculate the number of liters of water vapor produced when 25.0 liters of oxygen gas are consumed.

Answer 1

The volume of water vapor produced in the combustion of 25.0 liters of O2 for propane combustion is 20.0 liters of H2O following the stoichiometric ratio from the balanced chemical equation.

Step-by-step explanation:

The student is asking about the calculation of the volume of water vapor produced in the combustion of propane gas when a given volume of oxygen gas is consumed. The balanced chemical equation for the combustion of propane is
`C_3H_8(g) + 5O_2(g) - > 3CO_2(g) + 4H_2O(g).`

Given that 25.0 liters of O2 are consumed, we can use stoichiometry based on the balanced equation to find the volume of H2O produced. Since the reaction shows that 5 moles of O2 produce 4 moles of H2O, the volume of water vapor produced would be (4/5) × 25.0 liters = 20.0 liters of H2O.

Answer 2

Answer: 20L of H2O

Step-by-step explanation:

C3H8 + 5O2 → 3CO2 + 4H2O

Recall 1mole of a gas contains 22.4L at stp

5moles of O2 contains = 5 x 22.4 = 112L

4moles of H2O contains = 4 x 22.4 = 89.6L

From the equation,

112L of O2 produced 89.6L H2O

There for 25L of O2 will produce XL of H2O i.e

XL of H2O = (25 x 89.6)/112 = 20L