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 choices291 kJ191 kJ491 kJ391 kJ

First, we need to determine the number of moles of Barium (Ba) in 200 g.

Moles of Ba = mass / molar mass Moles of Ba = 200 g / 137.3 g/mol = 1.46 moles

The given reaction shows that 1 mole of Ba reacts to release 431 kJ of heat. Therefore, the amount of heat released by 1.46 moles of Ba would be:

Heat released = ΔH * moles of Ba Heat released = -431 kJ/mol * 1.46 mol = -629.26 kJ

However, the answer choices are all positive, suggesting that we are looking for the magnitude of the heat released, not the direction. Therefore, the heat released is 629.26 kJ.

None of the given answer choices match this result. There may be a mistake in the question or the answer choices.

First, we need to calculate the number of moles of Barium (Ba) using the given mass and molar mass.

Moles of Ba = mass / molar mass Moles of Ba = 200 g / 137.3 g/mol = 1.46 moles

The given reaction shows that 1 mole of Ba reacts to release 431 kJ of heat. Therefore, we can set up a proportion to find out how much heat is released when 1.46 moles of Ba react.

1 mole Ba : 431 kJ = 1.46 moles Ba : x kJ

Solving for x gives us:

x = 1.46 moles Ba * 431 kJ / 1 mole Ba = 629.06 kJ

However, none of the provided answer choices match this result. There may be a mistake in the problem or the answer choices.