Answer:
3.82 g.
Step-by-step explanation:
To calculate the mass of sulfuric acid produced from 1.25 g of sulfur, we'll have to balance the chemical equations and use stoichiometry.
Starting with the first reaction:
2S(s) + 3O2(g) → 2SO3(g)
Since 1.25 g of sulfur is reacted, the number of moles of sulfur can be calculated as:
moles = mass / molar mass = 1.25 g / 32 g/mol = 0.03906 mol
Next, using the mole ratio from the balanced chemical equation, we can find the number of moles of sulfur trioxide produced:
2 moles of sulfur produce 2 moles of sulfur trioxide, so:
moles of SO3 = moles of S x (moles of SO3 / moles of S) = 0.03906 mol x (2 mol / 2 mol) = 0.03906 mol
Finally, we move on to the second reaction, the dissolution of sulfur trioxide in water:
SO3(g) + H2O(l) → H2SO4(l)
Using the mole ratio from this balanced equation, we can find the number of moles of sulfuric acid produced:
1 mole of sulfur trioxide reacts with 1 mole of water to produce 1 mole of sulfuric acid, so:
moles of H2SO4 = moles of SO3 x (moles of H2SO4 / moles of SO3) = 0.03906 mol x (1 mol / 1 mol) = 0.03906 mol
The mass of sulfuric acid produced can be calculated using the moles and the molar mass of sulfuric acid:
mass = moles x molar mass = 0.03906 mol x 98 g/mol = 3.82 g
Therefore, if 1.25 g of sulfur is reacted, the mass of sulfuric acid produced is approximately 3.82 g.