Question

when 4.2 gram of sodium hydrogen carbonate is added to a solution of hydrochloric acid during 10.0 gram it is observed that 2.2 gram of carbon dioxide is released into the atmosphere the Residue left Behind is found to be 12.0 gram show that this observation are in agreement with the law of conversation of mass

Answer 1

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Molecular weights

NaHCO3 = 84

CH3COOH = 60

CH3COONa = 82

CO2 = 44

84 grams NaHCO3 reacts with 60 grams of CH3COOH to form 82 grams of CH3COONa and 44 grams of CO2.

Given 6.3 grams of NaHCO3 reacts with 15 grams of CH3COOH.

Number of moles of NaHCO3 = given weight/ moleculear weight

= 6.3/84

= 0.075moles

Number of moles of CH3COOH = 15/60

= 0.25

It is to be remember that 1 mole of reactants combines to produce 1 mole of products

Thus 0.075 moles of NaHCO3 will reacts with 0.075 moles of to produce 0.075 moles of CO2.

Weight of 1 mole of CO2 is = 44 grams

Weight 0.075 moles of CO2 = (0.075*44)

= 3.3 grams
So 3.3 grams of CO2 will be produced in the reaction.