What mass of water will be released during the combustion of 5.6 liters of propane?

Given:

V (C3H8) = 5.6 L;

To find:

m (H2O) -?

Decision:

1) We compose the reaction equation typical for the problem condition:

C3H8 + 5O2 – (heat) -> 3CO2 + 4H2O;

2) Find the amount of propane contained in 5.6 liters of gas:

n (C3H8) = V: Vm = 5.6 L: 22.4 L / mol = 0.25 mol;

3) We compose a logical expression:

if 1 mol of C3H8 gives 4 mol of H2O,

then 0.25 mol C3H8 will give x mol H2O,

then x = 1 mol.

4) Find the mass of water formed as a result of the reaction:

m (H2O) = n * M = 1 mol * 18 g / mol = 18 g;

Answer: m (H2O) = 18 g.