Final answer:
To find the grams of H2SO4 formed from 6L of SO3 at STP, one needs to write the balanced chemical equation, calculate the number of moles of SO3 based on its volume at STP, and then multiply by the molar mass of H2SO4. A total of 26.25 grams of H2SO4 would be produced.
Step-by-step explanation:
To calculate how many grams of H2SO4 will be formed by reacting 6 L of SO3 gas with sufficient water at STP, you would first need to write the balanced chemical equation for the reaction between sulfur trioxide (SO3) and water (H2O). The equation is:
SO3(g) + H2O(l) → H2SO4(aq)
Now, at STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. To find the number of moles of SO3, we divide the volume of SO3 by the molar volume of a gas at STP:
Moles of SO3 = 6 L / 22.4 L/mol = 0.2679 mol
The molar mass of H2SO4 is approximately 98 g/mol. Therefore, the mass of H2SO4 produced can be calculated by multiplying the moles of SO3 by the molar mass of H2SO4:
Mass of H2SO4 = 0.2679 mol × 98 g/mol = 26.25 g
Thus, 26.25 grams of H2SO4 would be formed.