What Determines the Limiting Reactant in a Chemical Reaction?

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SUMMARY

The limiting reactant in the reaction between sodium hydroxide (NaOH) and carbon dioxide (CO2) is NaOH, as determined by the stoichiometric calculations based on their molar masses. Given 400 g of NaOH and 308 g of CO2, the reaction produces 530 g of sodium carbonate (Na2CO3). After the reaction, 88 g of CO2 remains unreacted. These calculations utilize the molar masses of NaOH (47 g/mol) and CO2 (42 g/mol) to establish the limiting reactant and the resultant products.

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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
 
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Cursed said:
NaOH = 47. g/mol
CO2 = 42. g/mol

No & no.

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