# How many moles of Na2CO3 can be produced?

2NaOH+CO2-->Na2CO3+H20

Which reagent is the limiting reactant when 1.82 mol NaOH and 1.00 mol CO_2 are allowed to react?

NaOh is the limiting factor

my question:

How many moles of Na2CO3 can be produced?

thanks!

also how many grams of the excess reactant remain after the limiting factor is completely used? thanks so much guys!

@oli-thank you that's correct! how did you get the answer?

• 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.

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.

• Moles Of Na2co3

• 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 : 0.91

So 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.

• .91 moles

• 0.91 Moles.

• 0.91 mols