ΔH for the following reaction is -431 kJ. How much heat is released when 200 g of Ba(s) are reacted with an excess of H2O(l)? The molar mass of Barium is 137.3 g/mol. Ba(s) + 2 H2O(l) –> Ba(OH)2(aq) + H2(g) Group of answer choices191 kJ491 kJ391 kJ628 kJ
Question
ΔH for the following reaction is -431 kJ. How much heat is released when 200 g of Ba(s) are reacted with an excess of H2O(l)? The molar mass of Barium is 137.3 g/mol. Ba(s) + 2 H2O(l) –> Ba(OH)2(aq) + H2(g) Group of answer choices191 kJ491 kJ391 kJ628 kJ
Solution
First, we need to determine the number of moles of Barium (Ba) in 200 g.
Moles = mass / molar mass Moles = 200 g / 137.3 g/mol = 1.46 moles
The given reaction shows that the ΔH (heat change) of -431 kJ is for 1 mole of Ba reacting with H2O.
So, for 1.46 moles of Ba, the heat released would be:
Heat = moles * ΔH Heat = 1.46 moles * -431 kJ/mole = -629.06 kJ
Since heat is being released, we take the absolute value, so the heat released is 629.06 kJ.
Therefore, the closest answer is 628 kJ.
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