What Determines the Limiting Reactant in a Chemical Reaction?

[Cursed]

Homework Statement

Sodium hydroxide reacts with carbon dioxide as follows,

2 NaOH (s) + CO₂ (g) $$\rightarrow$$ Na₂CO₃ (s) + H₂O (g)​

Suppose 400. g of NaOH (s) is allowed to react with 308. g of CO₂ (g)

A. What is the limiting reactant? (Answer: NaOH)
B. What mass of sodium carbonate can be produced? (Answer: 530g)
C. What mass of excess reactant remains after the limiting reactant has been consumed? (Answer: 88g of CO₂ remains)

Homework Equations

N/A

The Attempt at a Solution

PART A

NaOH = 47. g/mol
CO₂ = 42. g/mol

(400. g) / (2 * 47. g/mol) = 4.3 mol NaOH
(308. g) / (42. g/mol) = 7.3 mol CO₂

Limiting Reactant: NaOH

PART B

Sodium carbonate (Na₂CO₃) = 120 g/mol

(120 g/mol Na₂CO₃/1 mol Na₂CO₃)(1 mol Na₂CO₃ / 2 mol NaOH)(1 mol NaOH / 47. g/mol NaOH)(400 g NaOH) = 425 g Na₂CO₃

 

[Borek]

NaOH = 47. g/mol
CO₂ = 42. g/mol

No & no.

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[Blueshift5]

Therefore, the maximum mass of sodium carbonate that can be produced is 425 g. However, since the limiting reactant is NaOH, the actual mass of sodium carbonate produced will be limited by the amount of NaOH available, which is 400 g.

PART C

After the reaction is complete, there will be 4.3 mol NaOH consumed. This leaves 7.3 mol CO2 remaining. Using the molar mass of CO2 (44 g/mol) and the number of moles remaining, we can calculate the mass of CO2 remaining.

(7.3 mol CO2) * (44 g/mol CO2) = 321.2 g CO2 remaining

Therefore, the mass of excess reactant remaining is 321.2 g - 308 g = 13.2 g CO2.