# How many grams of H2O and CO2 are formed by the reaction of 500 mol of citric acid? Here's the formula: ZnCO3 (s) + 2 C6H8O7 (aq) → Zn3(C6H5O7)2 (aq) + 3 H2O (l) + 3 CO2 (g) How many grams of H2O and CO2 are formed by the reaction of 500 mol of citric acid? According to given stoichiometry, for 2mols of citric acid 3 mols of H2O is formed and 3 mols of CO2 is formed.

Therefore for 1 mol of citric acid reacted,

H2O formed = 3/2 mol

CO2 formed = 3/2 mol

Therefore for 500 mol of citric acid,

H2O formed...

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According to given stoichiometry, for 2mols of citric acid 3 mols of H2O is formed and 3 mols of CO2 is formed.

Therefore for 1 mol of citric acid reacted,

H2O formed = 3/2 mol

CO2 formed = 3/2 mol

Therefore for 500 mol of citric acid,

H2O formed = 3/2 mol x 500 = 750 mol

CO2 formed = 3/2 mol x 500 = 750 mol

Therefore 750 mols of H2O and CO2 are produced.

Weight of H2O formed = 750 mol x 18 g per mol

= 13500 g.

Weight of CO2 formed = 750 mol x 44 g per mol

= 33000 g.

Approved by eNotes Editorial Team