Question

what is the number of grams of sodium acetate (NaC2H3O2) will you produce from 5.14 mole of sodium bicarbonate (NaHCO3)

Answer 1

Taking into account the reaction stoichiometry, 421.48 grams of NaC₂H₃O₂ are formed from 5.14 mole of sodium bicarbonate.

Reaction stoichiometry

In first place, the balanced reaction is:

HC₂H₃O₂ + NaHCO₃ → NaC₂H₃O₂ + CO₂ + H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• HC₂H₃O₂: 1 mole
• NaHCO₃: 1 mole
• NaC₂H₃O₂: 1 mole
• CO₂: 1 mole
• H₂O: 1 mole

The molar mass of the compounds is:

• HC₂H₃O₂: 60 g/mole
• NaHCO₃: 84 g/mole
• NaC₂H₃O₂: 82 g/mole
• CO₂: 44 g/mole
• H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• HC₂H₃O₂: 1 mole ×60 g/mole= 60 grams
• NaHCO₃: 1 mole ×84 g/mole= 84 grams
• NaC₂H₃O₂: 1 mole ×82 g/mole= 82 grams
• CO₂: 1 mole ×44 g/mole= 44 grams
• H₂O: 1 mole ×18 g/mole= 18 grams

Mass of sodium acetate formed

The following rule of three can be applied: if by reaction stoichiometry 1 mole of NaHCO₃ form 82 grams of NaC₂H₃O₂, 5.14 moles of  NaHCO₃ form how much mass of NaC₂H₃O₂?

$mass of NaC_{2} H_{3} O_{2} =\frac{5.14 moles of NaHCO_{3}x82 grams of NaC_{2} H_{3} O_{2}}{1 mole of NaHCO_{3}}$

mass of NaC₂H₃O₂= 421.48 grams

Then, 421.48 grams of NaC₂H₃O₂ are formed from 5.14 mole of sodium bicarbonate.

Learn more about the reaction stoichiometry:

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