1) Consider the formation of sulfur trioxide according to the reaction:
2 SO2(g) + O2(g) -> 2 SO3(g) delta H = -198
How much heat is evolved in the formation of 750 grams of SO3?
I am not sure how to start this problem.
193 x (750/2*molar mass SO3) = ?
So I did 193 X (750/2*80.0642) = 904.
The answer choices given are 10.6, 928, 1.86 X 10^3, 7.43 X 10^4, 3.71 X 10^3
And how did you get 193?
To find out how much heat is evolved in the formation of 750 grams of sulfur trioxide (SO3), you need to use the concept of stoichiometry and the given enthalpy change (ΔH) value.
1) Start by determining the molar mass of SO3. Sulfur has a molar mass of approximately 32 grams/mol, and oxygen has a molar mass of approximately 16 grams/mol. Therefore, the molar mass of SO3 is:
(32 g/mol of S) + (3 * 16 g/mol of O) = 80 g/mol of SO3
2) Next, calculate the number of moles of SO3 in 750 grams. To do this, divide the given mass by the molar mass of SO3:
Number of moles = Mass / Molar mass
Number of moles = 750 g / 80 g/mol ≈ 9.375 mol
3) Now, use the stoichiometry of the balanced chemical equation to find the heat evolved (ΔH) when 2 moles of SO2 react to form 2 moles of SO3. According to the given balanced equation, the ΔH of the reaction is -198 kJ.
So, for every 2 moles of SO3 formed, the heat evolved is -198 kJ.
4) Calculate the heat evolved for 9.375 moles of SO3. To do this, use a proportion:
Heat evolved = (Number of moles of SO3 × ΔH) / (2 moles of SO3)
Heat evolved = (9.375 mol × -198 kJ) / 2 mol
Simplifying this expression, you get:
Heat evolved = -927.75 kJ
Therefore, approximately -927.75 kJ of heat is evolved in the formation of 750 grams of SO3. Note that the negative sign indicates the release of heat during the reaction.