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
A. -0.4 kJ
B. 1776.7 kJ
C. 0.4 kJ
D. -1776.7 kJ
E. -6.2 kJ
The equation tells us that 285.8 kJ of heat are released per mol of liquid water formed. I would see how many mols are in 111.9 grams of water and that times 285.8 kJ/mol will tell me how many kJ of heat are released.
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 need to use the given delta Ho value and the molar mass of water.
1. Calculate the moles of H2O(l) formed:
Molar mass of H2O(l) = 2g/mol (from H2: 2g/mol + O: 16g/mol)
Moles of H2O(l) = mass / molar mass = 111.9g / 18g/mol = 6.2167 mol
2. Use the balanced equation and stoichiometry to find the moles of H2(g) used:
The balanced equation shows that 1 mole of H2(g) produces 1 mole of H2O(l).
Therefore, the moles of H2(g) used = 6.2167 mol.
3. Use the given delta Ho value to find the quantity of heat evolved:
Delta Ho = -285.8 kJ/mol
Heat evolved = Delta Ho * moles of H2(g) = -285.8 kJ/mol * 6.2167 mol = -1776.7 kJ.
Therefore, the correct answer is D. -1776.7 kJ.