Question

Which reagent is the limiting reactant when 2.22 mol NaOH and 1.20 mol CO2 are allowed to react?

Which reagent is the limiting reactant when 2.22   and 1.20   are allowed to react?

	NaOH
	CO2 Part B. How many moles of Na2CO3 can be produced? Part C. How many moles of the excess reactant remain after the completion of the reaction?

Answer 1

Solution :-

2 NaOH + CO2 ----- > Na2CO3 + H2O

2.22 mol NaOH and 1.20 mol CO2

Lets calculate the moles of the CO2 needed to react with 2.22 mol NaOH

2.22 mol NaOH * 1 mol CO2 / 2 mol NaOH = 1.11 mol CO2

So the moles of CO2 present are more than required

So NaOH is the limiting reactant

Part B) Now lets calculate the moles of the Na2CO3 that can be formed

2.22 mol NaOH * 1 mol Na2CO3 / 2 mol NaOH = 1.11 mol Na2CO3

Part C ) moles of excess reactant remain

CO2 is the excess reactant

So moles of CO2 remain = moles of CO2 total - moles of CO2 needed for the reaction

                                            = 1.20 mol – 1.11 mol

                                            = 0.09 mol CO2

So 0.09 mol CO2 will remain.