2h2s g 3o2 g 2so2 g 2h2o g H=-1036KG/MOL
calculate enthalpy change in kj when 18.2 g of hydrogen sulfide are burned
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To calculate the enthalpy change in kJ when 18.2 g of hydrogen sulfide (H2S) is burned, we need to use the given molar mass of H2S to determine the number of moles consumed.
1. Calculate the molar mass of H2S:
- The molar mass of hydrogen (H) is approximately 1 g/mol.
- The molar mass of sulfur (S) is approximately 32 g/mol.
- Therefore, the molar mass of H2S is (2 x 1 g/mol) + (1 x 32 g/mol) = 34 g/mol.
2. Calculate the number of moles of H2S:
- The given mass of H2S is 18.2 g.
- Using the molar mass of H2S (34 g/mol), we can calculate the number of moles: moles = mass / molar mass
- moles = 18.2 g / 34 g/mol = 0.535 mol (rounded to 3 decimal places).
3. Use the balanced chemical equation and the enthalpy change for combustion to calculate the enthalpy change for the given amount of H2S burned:
- From the balanced equation: 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
- The stoichiometric coefficient of H2S is 2, which means 2 moles of H2S react to produce 2 moles of H2O(g)
- According to the given enthalpy change: ΔH = -1036 kJ/mol
- Therefore, the enthalpy change for 0.535 mol of H2S burned is: ΔH = 0.535 mol x (-1036 kJ/mol).
4. Calculate the enthalpy change in kJ:
- ΔH = 0.535 x -1036 kJ = -553.16 kJ (rounded to 2 decimal places).
Therefore, the enthalpy change when 18.2 g of hydrogen sulfide is burned is approximately -553.16 kJ.
To calculate the enthalpy change when 18.2 g of hydrogen sulfide (H2S) is burned, we first need to determine the amount of H2S in moles.
1. Find the molar mass of H2S:
- Hydrogen (H) has an atomic mass of 1 g/mol. Since there are two hydrogen atoms in H2S, the contribution of hydrogen is 2 g/mol.
- Sulfur (S) has an atomic mass of 32 g/mol.
- The total molar mass of H2S is 2 + 32 = 34 g/mol.
2. Calculate the number of moles of H2S:
- Divide the given mass by the molar mass:
moles = mass / molar mass
moles = 18.2 g / 34 g/mol
3. Determine the balanced chemical equation for the reaction:
2H2S (g) + 3O2 (g) → 2SO2 (g) + 2H2O (g)
4. Use stoichiometry to find the number of moles of O2 required and the number of moles of products formed:
- According to the balanced equation, two moles of H2S react with three moles of O2.
- So, the number of moles of O2 required is (2/2) × (3/2) = 3/2 moles.
- The number of moles of products formed will be equal to the number of moles of reactants.
5. Calculate the enthalpy change using the given heat of formation (∆H) for H2S:
- The enthalpy change for the given reaction is stated as -1036 kJ/mol.
- Since 2 moles of H2S are required to produce this amount of heat, divide the given value by 2:
∆H = -1036 kJ/mol / 2
6. Calculate the enthalpy change for the given mass of H2S:
- Multiply the moles of H2S from step 2 by the enthalpy change (∆H) from step 5:
enthalpy change = moles × ∆H
Substituting the values calculated above will give you the enthalpy change in kJ when 18.2 g of H2S is burned.
I STRONGLY object to posts with no arrows, no plus signs, no caps (except in the wrong place), no periods, no parentheses, and no sentence structure. If you want us to help, you need to make your posts readable. I assume you intended to write
2H2S(g) + 3O2(g) ==> 2SO2(g) + 2H2O(g) delta H=-1036 kg/mol which should be kJ/mol.
1036 kJ/mol x (18.2g/molar mass H2S) = ?