Answer to Question #144623 in General Chemistry for Olivia

Question

Sodium hydroxide reacts with carbon dioxide according to the equation: NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l). a) Which is the "limiting reagent" when 7.4 g of NaOH and 0.44 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 0.90 g of Na2CO3 are obtained?

Answer 1

2NaOH_(s)+CO_2(g)=Na₂CO_3(s)+H₂O_(l)

Moles of NaOH in 7.4g

Moles=Mass/RMM

=7.4g/40gmol^-1

=0.185moles

Moles of CO₂ in 0.44g =0.44g/44gmol^-1

=0.01moles

If 0.01 moles of CO₂ were to react then, 2(0.01) moles NaOH was to be consumed

=0.02 moles of NaOH

If 0.185 moles of NaOH reacted then, 0.185moles/2 CO₂ reacted

=0.0925moles of CO₂

Since the number of moles of CO₂ were less than 0.0925 moles, it means CO₂ was the limiting reagent as all of it, CO₂, reacted while (0.185-0.02)g = 0.165g of NaOH remained in excess

Theoretical mass of Na₂CO₃ that was to be formed

Theoretical Moles of Na₂CO₃= 0.01moles

mass=0.01g*106gmol^-1

=1.06g

%yield of Na₂CO₃= (0.90g/1.06g)*100%

=84.91%