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Limiting Reactant in Chemistry: Identification and Calculation, Study notes of Chemistry

Chemistry

A portion of a chemistry textbook discussing the concept of limiting reactant in chemical reactions. It explains how to identify the limiting reactant by comparing the actual and theoretical mole ratios or calculating the number of moles obtained from each reactant. The document also covers the yield concept and its calculation.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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Chemistry101 Chapter3
22
LIMITINGREACTANT
· When 2 or more reactants are combined in non-stoichiometric ratios, the amount of
product produced is limited by the reactant that is not in excess (limiting
reactant).
Analogy:
The number of sundaes possible is limited by the amount of syrup, the limiting reactant.
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Download Limiting Reactant in Chemistry: Identification and Calculation and more Study notes Chemistry in PDF only on Docsity!

LIMITING REACTANT

∑ When 2 or more reactants are combined in non-stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess ( limiting reactant).

Analogy:

The number of sundaes possible is limited by the amount of syrup, the limiting reactant.

Limiting Reactant (Reagent) Problems always involve 2 steps:

  1. Identify the Limiting Reactant (LR) convert all masses to moles compare actual mole ratio to mole ratio given by the the balanced chemical equation

OR calculate the number of moles obtained from each reactant in turn. The reactant that gives the smaller amount of product is the Limiting R.eactant.

  1. Calculate the amount of product obtained from the Limiting Reactant

Example 1 Sodium hydrogen carbonate is prepared from NaCl and ammonium hydrogen carbonate, according to the equation:

NH 4 HCO 3 (aq) + NaCl(aq) NaHCO 3 (aq) + NH 4 Cl(aq)

If 0.300 moles of NH 4 HCO 3 are reacted with 0.2567 moles of NaCl, how many grams of NaHCO 3 are obtained?

1 NH 4 HCO 3 (aq) + 1 NaCl(aq) 1 NaHCO 3 (aq) + 1 NH 4 Cl(aq)

0.300 moles 0.2567 moles? g

L.R.

1 mole NaHCO 3 84.01 g NaHCO 3 ? g NaHCO 3 = 0.2567 moles NaCl x æææææææ x ææææææ 1 mole NaCl 1 mole NaHCO 3

= 21.57 g NaHCO 3

Solution recommended by textbook:

1 Mg(s) + 2 HCl(aq) 1 MgCl 2 (aq) + 1 H 2 (g)

1 mole 1 mole

1.4 g x æææ 8.1 g x æææ

24.31 g 36.46 g

0.0576 moles 0.222 moles

Calculate the number of moles obtained from each reactant in turn. The reactant that gives the smaller amount of product is the Limiting Reactant.

1 mole H 2 2.02 g H 2

? g H 2 = 0.222 moles HCl x æææææ x æææææ = 0.22 g H 2

2 moles HCl 1 mole H 2

1 mole H 2 2.02 g H 2

? g H 2 = 0.0576 moles Mg x æææææ x æææææ = 0.12 g H 2

1 mole Mg 1 mole H 2 smaller!

(correct answer)

Since Mg produces the smaller amount of product, Mg is the L.R.

THE YIELD CONCEPT

∑ Quantities of product calculated represent the maximum amount obtainable (100 % yield)

∑ Most chemical reactions do not give 100 % yield of product because of:

side reactions (unwanted reactions)

reversible reactions ( reactants products)

losses in handling and transferring

Actual Yield Percent Yield = ææææææææ x 100 Theoretical Yield

Actual Yield: Amount of product actually obtained (experimental)

Theoretical Yield: Maximum amount of product obtainable (calculated from equation)

Example 2 Sodium hydrogen carbonate is prepared from NaCl and ammonium hydrogen carbonate, according to the equation:

NH 4 HCO 3 (aq) + NaCl(aq) NaHCO 3 (aq) + NH 4 Cl(aq)

If 0.300 moles of NH 4 HCO 3 are reacted with 0.2567 moles of NaCl, and 10.45 g of NaHCO 3 are obtained, what is the percent yield?

  1. First calculate the maximum amount obtainable (theoretical yield) from the given quantities (theoretical yield)

1 NH 4 HCO 3 (aq) + 1 NaCl(aq) 1 NaHCO 3 (aq) + 1 NH 4 Cl(aq)

0.300 moles 0.2567 moles? g

L.R.

1 mole NaHCO 3 84.01 g NaHCO 3

? g NaHCO 3 = 0.2567 moles NaCl x æææææææ x ææææææ = 21.57 g NaHCO 3 (theoretical yield)

1 mole NaCl 1 mole NaHCO 3

  1. Second, calculate % yield from actual and theoretical yield

Actual Yield 10.45 g Percent Yield = ææææææææ x 100 = æææ x 100 = 48.45 % Theoretical Yield 21.57 g