Final answer:
To find the amount of H2SO4 formed by the reaction of 6L of SO3 gas with sufficient water at STP, we need to use stoichiometry. The balanced equation is SO3 + H2O -> H2SO4. Use the molar volume at STP (22.4 L/mol) to convert the volume of SO3 to moles, then convert moles to grams by multiplying by the molar mass of H2SO4.
Step-by-step explanation:
The question asks how many grams of H2SO4 will be formed by the reaction of 6L of SO3 gas with sufficient water at STP. To solve this problem, we need to use stoichiometry. The balanced equation for the reaction is:
SO3 + H2O → H2SO4
According to the balanced equation, for every 1 mole of SO3 reacted, 1 mole of H2SO4 is formed. We can use the molar volume at STP (22.4 L/mol) to convert the volume of SO3 to moles:
6 L SO3 * (1 mol SO3 / 22.4 L SO3) = 0.268 mol SO3
Since the reaction is 1:1, the number of moles of H2SO4 formed will also be 0.268 mol. To convert moles to grams, we need to multiply by the molar mass of H2SO4 (98.09 g/mol):
0.268 mol * 98.09 g/mol = 26.3 g H2SO4