YouCalculate the heat released when 74.6 grams of SO2 reacts according to the following equation?2SO2+O2-->2SO3 deltaH=-99.1 kJ
Question
YouCalculate the heat released when 74.6 grams of SO2 reacts according to the following equation?2SO2+O2-->2SO3 deltaH=-99.1 kJ
Solution
Sure, here's how you can calculate the heat released:
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First, we need to know the molar mass of SO2. The molar mass of S (Sulfur) is approximately 32.06 g/mol and that of O2 (Oxygen) is 16.00 g/mol. Therefore, the molar mass of SO2 (Sulfur dioxide) is 32.06 g/mol + 2(16.00 g/mol) = 64.06 g/mol.
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Next, we calculate the number of moles of SO2 in 74.6 grams. We use the formula: Number of moles = mass / molar mass So, the number of moles of SO2 = 74.6 g / 64.06 g/mol = 1.165 mol.
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The balanced chemical equation tells us that 2 moles of SO2 will release 99.1 kJ of heat. Therefore, 1 mole of SO2 will release 99.1 kJ / 2 = 49.55 kJ of heat.
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Finally, we can find out the heat released by 1.165 moles of SO2 by multiplying the heat released by one mole with the number of moles. So, the heat released = 1.165 mol * 49.55 kJ/mol = 57.7 kJ.
Therefore, 74.6 grams of SO2 will release approximately 57.7 kJ of heat when it reacts according to the given equation.
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