Question

2FeCl + 3Na2CO2= Fe2 (CO3)3 + 6NaCl
If you begin a reaction with 127.490 g of Na2CO3, how many moles of NaCl can you theoretically produce, assuming an excess of FeCl3 is present?
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Answer 1

Taking into account the reaction stoichiometry, 2.405 moles of NaCl are formed when  127.49 grams of Na₂CO₂ reacts with excess of FeCl₃.

Reaction stoichiometry

In first place, the balanced reaction is:

2 FeCl+ 3 Na₂CO₂ → Fe₂(CO₃)₃ + 6 NaCl

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:

• FeCl: 2 moles
• Na₂CO₂: 3 moles
• Fe₂(CO₃)₃: 1 mole
• NaCl: 6 moles

The molar mass of the compounds is:

• FeCl: 91.3 g/mole
• Na₂CO₂: 106 g/mole
• Fe₂(CO₃)₃: 291.7 g/mole
• NaCl: 58.45 g/mole

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

• FeCl: 2 moles ×91.3 g/mole= 182.6 grams
• Na₂CO₂: 3 moles ×106 g/mole= 318 grams
• Fe₂(CO₃)₃: 1 mole ×291.7 g/mole= 291.7 grams
• NaCl: 6 moles ×58.45 g/mole= 350.7 grams

Mass of NaCl produced

The following rule of three can be applied: if by reaction stoichiometry 318 grams of Na₂CO₂ form 6 moles of NaCl, 127.49 grams of Na₂CO₂ form how many moles of NaCl?

$moles of NaCl=\frac{127.49 grams of Na_{2} CO_{2} x6 moles of NaCl }{318grams of Na_{2} CO_{2} }$

moles of NaCl= 2.405 moles

Then, 2.405 moles of NaCl are formed when  127.49 grams of Na₂CO₂ reacts with excess of FeCl₃.

Learn more about the reaction stoichiometry:

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