For the reaction
H2+1/2 O2 -> H2O ΔH°=286 kJ/mol
Calculate the enthalpy change when 2.44 g of water is produced.
a. 117 kJ
b. 698 kJ
c. –698 kJ
d. 38.7 kJ
e. –38.7 kJ
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First, we need to calculate the number of moles of water produced.
The molar mass of water (H2O) is 18.015 g/mol.
Number of moles of water = mass of water (g) / molar mass of water
Number of moles of water = 2.44 g / 18.015 g/mol
Number of moles of water = 0.1356 mol
Now, we can use the molar ratio between water and the reaction to calculate the enthalpy change.
From the balanced equation, we can see that the molar ratio between water and the reaction is 1:1. This means that for every 1 mole of water produced, the enthalpy change is 286 kJ.
Therefore, for 0.1356 mol of water produced, the enthalpy change is:
Enthalpy change = 0.1356 mol water * 286 kJ/mol
Enthalpy change = 38.7 kJ
The enthalpy change when 2.44 g of water is produced is 38.7 kJ.
Therefore, the correct answer is d. 38.7 kJ.
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