How many moles of Na2CO3 can be produced?
How many moles of Na2CO3 can be produced from the reaction between sodium hydroxide (NaOH) and carbon dioxide (CO2)? The balanced chemical equation for the reaction is:
2NaOH + CO2 → Na2CO3 + H2O
Given that 1.82 moles of NaOH and 1.00 mole of CO2 are allowed to react, which reagent is the limiting reactant? I believe NaOH is the limiting factor. Based on this information, how many moles of Na2CO3 can be produced? Thank you!
7 Answers
Nov 10, 2024
Of course you want to know how to get this answer:
You have identified the NaOH as the limiting reactant
The amount of Na2CO3 is governed by the mol of the limiting reactant
Look at the balanced equation:
2mol NaOH will produce 1 mol Na2CO2
Note : Mol Na2CO3 is half mol of NaOH
You have 1.82 mol NaOH
Therefore you have 1.82/2 = 0.91 mol Na2CO3 produced.
Your additional question:
From the equation:
1.82 mol NaOH will react with 0.91 mol of CO2
You have 1.00 mol CO2.
Mol CO2 unreacted = 1.00- 0.91 = 0.09 mol CO2
Molar mass CO2 = 44g/mol
0.09 mol CO2 = 0.09*44 = 3.96g CO2 is left unreacted.
If NaOH is the limiting factor, then that means the maximum amount of NaOH that can react will be 1.82 mol, and this reacts with half the moles of Carbon which is 0.91 Moles.
From the equation; the molar ratio is 2NaOH:Na2CO3
= 2 :1 =1.82 : 0.91So 0.91 moles of Na2CO3 is the maximum that can be produced.
So we know 0.91 moles of CO2 react with 1.82 moles of NaOH. This means that 0.91(162+12)
grams of CO2 will react (Moles*Mr = mass)
Which is 40.04 g of CO2 will react.
From the compositions at the beginning, you put in 1 mol of CO2 which is 1(12+162) = 44g
So the amount left is (44-40.04 = 3.96 g)
Sorry if I have made any mistakes or given any incorrect information. I have a very basic understanding of chemistry.
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Limiting Reagent : NaOH 1.14 moles Na2CO3 0.06 moles of Excess Reactant left after completion.
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