The value of ΔH° for the reaction below is - 790 kJ.2S(s) + 3O2(g) → 2SO3(g)The enthalpy change accompanying the reaction of 0.95 g of S is ________kJ.Question 4Select one:a.23b.-23c.-12d.12e.-790
Question
The value of ΔH° for the reaction below is - 790 kJ.2S(s) + 3O2(g) → 2SO3(g)The enthalpy change accompanying the reaction of 0.95 g of S is ________kJ.Question 4Select one:a.23b.-23c.-12d.12e.-790
Solution
To solve this problem, we need to first understand that ΔH° given is for the reaction of 2 moles of S(s).
- Calculate the molar mass of S(s). The molar mass of S(s) is approximately 32.06 g/mol.
- Calculate the number of moles in 0.95 g of S(s). This is done by dividing the mass of S(s) by its molar mass. So, 0.95 g / 32.06 g/mol = 0.0296 mol.
- Since the given ΔH° is for 2 moles of S(s), we need to find the ΔH° for 0.0296 mol. This is done by multiplying the given ΔH° by the ratio of 0.0296 mol to 2 mol. So, -790 kJ * (0.0296 mol / 2 mol) = -11.69 kJ.
- Round to the nearest whole number, the enthalpy change accompanying the reaction of 0.95 g of S is -12 kJ.
So, the correct answer is c. -12.
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