For the reaction so3 + h2o h2so4, calculate the percent yield if 500. g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid.

The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula

% yield = actual yield / theoretical yield x100

actual yield = 575 grams

to calculate theoretical yield

find the moles of SO3 used = mass/molar mass

= 500g / 80 g/mol = 6.25 moles

SO3+H2O=H2SO4

by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also = 6.25 moles

the theoretical yield of H2SO4 is therefore = moles / molar mass

= 6.25 x98 = 612.5 grams

%yield is therefore = 575 g/612 g x100 = 93.9 %