Consider the following reaction between sulfur trioxide and water: SO3(g)+H2O(l)=H2SO4(aq) A chemist allows 61.5 g fo...

Consider the following reaction between sulfur trioxide and water:

SO3(g)+H2O(l)=H2SO4(aq)

A chemist allows 61.5 g fo SO3 and 11.2 g of H2O to react. When the reaction is finished, the chemist collects 51.0g of H2SO4.

1) Determine the theoretical yield for the reaction.

2) Determine the percent yield for the reaction.

Solutions

Expert Solution

SO_3(g) + H₂O_(l) = H₂SO_4(aq)

Moles 61.5/80 11.2/18

= 0.76875 = 0.622

Water is limiting reagnet so all caluclations should be according to water

expected moles of H₂SO_4(aq) formed = 0.622 moles

Mass of H₂SO_4(aq) formed = 0.622x98 = 60.9 75 gram ( theoretical yield)

% yield = (actual mass formed/expected mass )x100 = ( 51.0g/60.9 75)x100 = 83.64 %

queen_honey_blossom answered 10 months ago

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