A sample of gas contains 0.1500 mol of CH4(g) and 0.1500 mol of H2O(g) and occupies a volume of 13.0 L. The following reaction takes place:
CH_(g) + H2O(g) 3H2(g) + CO(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
L

26.0L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of CH4 reacts with 1 mole of H2O to produce 1 mole of CO and 3 moles of H2.

That is, 1 mole of each reactant produce 4 moles of gases

As in the reaction, 0.1500 moles of CH4 and 0.1500 moles of H2O are added, 0.1500 moles of CO and 0.4500 moles of H2 are produced.

Before the reaction, the moles of gas are 0.3000 moles and after the reaction the moles are 0.6000 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

Where V is volume and n are moles of 1, initial state and 2, final state.

Replacing:

V1 = 13.0L

n1 = 0.3000 moles

V2 = ?

n2 = 0.6000 moles

13.0L*0.6000 moles / 0.3000 moles = V2

V2 = 26.0L is the volume of the sample after the reaction

2b2t

explanation:

2b2t