Question

Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 9.90 mL of water (d= 1.00 g/mL) and 22.7 g of SO3? How much of the reagent in excess is left over?

Answer 1

molar mass of SO3 = 80 g/mol
number of mole of SO3 = (given mass)/(molar mass)
= 22.7/80
= 0.28 mole
mass of H2O = (volume)*(density)
= (1*9.90) g
= 9.90 g
molar mass of H2O = 18 g/mol
number of mole of H2O = 9.90/18
= 0.55 mole

reaction taking place
SO3 + H2O --> H2SO4
according to reaction given
1 mole of SO3 required 1 mole of H2O
0.28 mole of SO3 required 0.28 mole of H2O
but we have 0.55 mole of H2O
so, H2O is in excess
and SO3 is limiting reagent

so,
number of mole of H2O remaing = 0.55-0.28
= 0.27 mole
mass of H2O remaing = (number of mole of H2O remins)*(molar mass)
= 0.27*18
= 4.86 g

again according to reaaction
1 mole of SO3 give 1 mole of H2SO4
0.28 mole of SO3 give 0.28 mole of H2SO4
number of mole of H2SO4 formed = 0.28 mole
mass of H2SO4 formed = (number of mole of H2SO4 formed)*(molar mass)
= 0.28*98
= 27.44 g