Question

Consider the reaction:

H₂g)+1/2O₂ (g)→H₂O(l); ΔH = -285.84 kJ/mol.
How many grams of hydrogen gas are needed to produce 975 kJ of heat?

Answer 1

To produce 975 kJ of heat, 6.87 grams of hydrogen gas are required, based on the enthalpy change of -285.84 kJ/mol for the formation of liquid water from hydrogen and oxygen gases.

Step-by-step explanation:

To determine how many grams of hydrogen gas are needed to produce 975 kJ of heat, we use the provided enthalpy change (ΔH) of the reaction.

Given the reaction H2(g) + 1/2O2(g) → H2O(l), the ΔH is -285.84 kJ/mol, meaning that 285.84 kJ of heat is released when 1 mole of hydrogen gas reacts with 1/2 mole of oxygen gas to form 1 mole of liquid water.

Since enthalpy change is an extensive property, it is directly proportional to the amount of the reactant consumed.

To produce 975 kJ, we first calculate the moles of hydrogen gas required using the enthalpy change:

975 kJ * (1 mol H2 / 285.84 kJ) = 3.41 mol H2

With the molar mass of hydrogen gas (2.016 g/mol), we can find the mass:

3.41 mol H2 * 2.016 g/mol = 6.87 g H2

So 6.87 grams of hydrogen gas are needed to produce 975 kJ of heat.