For the balanced equation shown below, if the reaction of 40.8 grams of C6H6O3 produces a 42.8% yield, how many grams of H2O would be produced ? C6H6O3+6O2=>6CO2+3H2O

Mass of H₂O produced = 7.488 g

Given

40.8 g of C₆H₆O₃

42.8% yield

Required

mass of H₂O

Solution

Reaction

C₆H₆O₃+6O₂=>6CO₂+3H₂O

mol of C₆H₆O₃(MW = 126 g/mol) :

= mass : MW

= 40.8 g : 126 g/mol

= 0.324

Mol H₂O :

= 3/1 x mol C₆H₆O₃

= 3/1 x 0.324

= 0.972

Mass H₂O(theoretical) :

= 0.972 x 18 g/mol

= 17.496 g

%yield = (actual/theoretical) x 100%

0.428 = actual / 17.496

actual = 7.488 g

it would be d. henry moseley

hope this !

and sorry if i'm late. have a good rest of your day! : d

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