Asked by Sean
If the percent yeild of the following equation is 55%, how many grams of H2SO4 are produced when 4.88 x 10^24 molecules of SO3 are combined with excess water?
SO3(g)+H2O(l)->H2SO4(aq).
what would be the formula to solve this equation
SO3(g)+H2O(l)->H2SO4(aq).
what would be the formula to solve this equation
Answers
Answered by
DrBob222
SO3 + H2O --> H2SO4
mols SO3 = 4.88E24/6.02E23 = approx 8 but you need more than an estimate.
Convert mols SO3 to mols H2SO4 using the coefficients in the balanced equation. Since 1 mol SO3 produces 1 mol H2SO4, then mols SO3 = mols H2SO4.
Convert mols H2SO4 to grams.
g H2SO4 = mols H2SO4 x molar mass H2SO4 = approx 800 g and this is if the reaction is 100%. It isn't so
approx 800 x 0.55 = ?
mols SO3 = 4.88E24/6.02E23 = approx 8 but you need more than an estimate.
Convert mols SO3 to mols H2SO4 using the coefficients in the balanced equation. Since 1 mol SO3 produces 1 mol H2SO4, then mols SO3 = mols H2SO4.
Convert mols H2SO4 to grams.
g H2SO4 = mols H2SO4 x molar mass H2SO4 = approx 800 g and this is if the reaction is 100%. It isn't so
approx 800 x 0.55 = ?
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