For the reaction So3 + H2O -> H2SO4, how many grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid
1 answer:
Answer:
320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.
Explanation:
The balanced reaction is:
SO₃ + H₂O → H₂SO₄
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- SO₃: 1 mole
- H₂O: 1 mole
- H₂SO₄: 1 mole
Being the molar mass of each compound:
- SO₃: 80 g/mole
- H₂O: 18 g/mole
- H₂SO₄: 98 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- SO₃: 1 mole* 80 g/mole= 80 grams
- H₂O: 1 mole* 18 g/mole= 18 grams
- H₂SO₄: 1 mole* 98 g/mole= 98 grams
Then you can apply the following rule of three: if 1 mole of sulfuric acid is produced by the reaction of 80 grams of sulfur trioxide, 4 moles of sulfuric acid is produced from how much mass of sulfur trioxide?
mass of sulfur trioxide= 320 grams
<u><em>320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.</em></u>
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Answer:
Total pressure at equilibrium is 0.2798atm.
Explanation:
For the reaction:
H₂S(g) ⇄ H₂(g) + S(g)
Kp is defined as:
If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:
H₂S(g): 0.150atm - x
H₂(g): x
S(g): x
Replacing in Kp:
X² = 0.1251 - 0.834X
X² + 0.834X - 0.1251 = 0
Solving for X:
X = -0.964 → False solution: There is no negative pressures
X = 0.1298
Thus, pressures are:
H₂S(g): 0.150atm - 0.1298atm = <em>0.0202atm</em>
H₂(g): <em>0.1298atm</em>
S(g): <em>0.1298atm</em>
Thus, total pressure in the container at equilibrium is:
0.0202atm + 0.1298atm + 0.1298atm = <em>0.2798atm</em>