1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →
Na2CO3(s) + H₂O(1).
a) Which is the limiting reagent when 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react?
b) How many grams of Na₂CO3 are expected to be produced ?
c) What would be the percent yield if 6.0 g of Na2CO3 are obtained?
One dimensional analysis

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

As 80 g NaOH produces 106 g of Na₂CO₃.

Therefore 5 g NaoH will produce ;

106 / 80 x 5 = 6.625 g

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