SO3 + H2O = H2SO4
If you react 34 g of water and have excess SO3, how much H2SO4 would you theoretically produce
how many moles of water do you have? 1? then you get one mole of sulfuric acid, "theoretically"
SO3 + H2O = OH + H2SO3
To calculate the amount of H2SO4 produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the substance that gets completely consumed and determines the maximum amount of product that can be formed.
Let's begin by calculating the molar mass of water (H2O):
H: 2 x 1.007 g/mol = 2.014 g/mol
O: 1 x 16.00 g/mol = 16.00 g/mol
Total molar mass of H2O = 2.014 + 16.00 = 18.014 g/mol
Next, we need to convert the mass of water to moles:
34 g H2O * (1 mol H2O / 18.014 g H2O) = 1.88 mol H2O
The balanced chemical equation tells us that 1 mole of SO3 reacts with 1 mole of water to produce 1 mole of H2SO4. Since there is excess SO3, we know that all the water will react completely.
Therefore, the theoretical yield of H2SO4 will be equal to the amount of H2O used, which is 1.88 mol.
Finally, we can convert the moles of H2SO4 to grams:
1.88 mol H2SO4 * (98.09 g H2SO4 / 1 mol H2SO4) = 184.5 g H2SO4
Therefore, theoretically, you would produce 184.5 grams of H2SO4 by reacting 34 grams of water with excess SO3.