What is the quantity of heat evolved when 111.9 grams of H2O(l) is formed from the combustion of H2(g) and O2(g)?

H2(g) + 1/2 O2(g) --> H2O(l) Delta Ho = -285.8 kJ

See response above.

To find the quantity of heat evolved when 111.9 grams of H2O(l) is formed from the combustion of H2(g) and O2(g), we can calculate the heat of reaction using the given enthalpy change.

The equation provided is:
H2(g) + 1/2 O2(g) --> H2O(l) ∆H° = -285.8 kJ

First, we need to calculate the number of moles of H2O(l) produced using the given mass of 111.9 grams and the molar mass of water (H2O), which is approximately 18.0153 g/mol.

Number of moles = Mass / Molar mass
Number of moles of H2O(l) = 111.9 g / 18.0153 g/mol

Once we have the number of moles of H2O(l), we can use the stoichiometry of the balanced equation to determine the molar ratio between H2O(l) and the given reaction.

From the balanced equation, the stoichiometric ratio of H2(g) to H2O(l) is 1:1. This means that for every 1 mole of H2(g) reacted, 1 mole of H2O(l) is produced.

Since the stoichiometric ratio is 1:1, the number of moles of H2(g) reacted will be the same as the number of moles of H2O(l) produced.

Now, we can calculate the heat evolved using the formula:

Heat evolved = ∆H° × moles of H2O(l) produced

Substitute the values into the formula:

Heat evolved = -285.8 kJ/mol × (moles of H2O(l) produced)

Now, calculate the number of moles and substitute it into the formula to find the quantity of heat evolved.