Limiting Reactants, Reaction Yield, and Solution Concentration in Chemistry

Limiting Reactant

The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.

Percent Yield

Actual yield is often less than the theoretical yield. Percent yield is calculated using the formula: actual yield = %yield/100% x Theoretical yield.

Balanced Equation for Ammonia Reaction

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Molar Mass of NH3

17 g/mol

Molar Mass of O2

32 g/mol

Moles of NH3 from 2 g

2 g / 17 g/mol = 0.118 mol of NH3

Moles of O2 from 4 g

4 g / 32 g/mol = 0.125 mol of O2

Amount of NO produced

0.118 mol of NH3 produces 0.118 mol of NO.

Amount of H2O produced

0.1 mol of O2 produces 0.15 mol of H2O (6/4 x 0.1 mol).

Excess NH3 remaining

0.018 mol of NH3 remains after the reaction.

Reaction of SiO2 and Carbon

100 grams of SiO2 react with an excess of powdered carbon to form 51.4 grams of SiC.

Limiting Reagent in Water Formation

The reactant that produces the lesser amount of product is the limiting reagent.

Reaction of Silicon Dioxide and Hydrogen Fluoride

SiO2 + 4HF → SiF4 + 2H2O

Molar Mass of SiO2

60 g/mol

Molar Mass of HF

20 g/mol

Moles of SiO2 from 30 g

30 g of SiO2 is 30 g / 60 g/mol = 0.5 mol of SiO2.

Moles of HF from 30 g

30 g of HF is 30 g / 20 g/mol = 1.5 mol of HF.

Grilling Dilemma

At Costco, you can only buy 8 buns, 15 patties, and 44 cheese slices.

Moles of Silicon Tetrafluoride

How many moles of silicon tetrafluoride would be formed from the reaction between 4.5 mol of silicon dioxide and 6.0 mol of hydrogen fluoride?

Hydrogen and Oxygen Reaction

2H2(g) + O2(g) → 2H2O(g)

Molar mass of SiC

40.1 g/mol

Theoretical yield of SiC

1.66 mol → 1.66 x 40.1 = 66.7 g

Actual yield of SiC

51.4 g

% yield

% yield = (51.4/66.7) x 100% = 77.1%

Reaction (I)

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Reaction (II)

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Moles of SiO2

100/60.1 = 1.66 mol

Left over N2

0.3 moles

Left over NH3

1.4 moles

Moles of NH3 produced from N2

2.0 moles N2 x (2/1) = 4.0 moles NH3

Moles of NH3 produced from H2

5.1 moles H2 x (2/3) = 3.4 moles NH3

Moles of NO produced from NH3

3.4 moles NH3 x (4/4) = 3.4 moles NO

Moles of NO produced from O2

2.5 moles O2 x (4/5) = 2.0 moles NO

Moles of H2O produced from O2

2.5 moles O2 x (6/5) = 3.0 moles H2O

Limiting reactant in Reaction (I)

H2 is the limiting reactant.

Limiting reactant in Reaction (II)

O2 is the limiting reactant.

Q1

What are the amounts of NO and H2O produced?

Q2

What else remains after the reaction?

Solutions

Homogeneous mixtures composed of two or more substances uniformly dispersed at the molecular or atomic level.

Solvent

The substance present in larger quantity that dissolves the solute.

Solute

The substance present in smaller quantity that is dissolved in the solvent.

Solubility

The ability of a substance (solute) to dissolve in a solvent to form a solution, influenced by factors such as temperature, pressure, and the nature of the solvent and solute.

Concentration

Describes the amount of solute dissolved in a given amount of solvent or solution, expressed in various units.

Molarity

The number of moles of solute per liter of solution.

Molality

The number of moles of solute per kilogram of solvent.

Mass percent

The mass of solute divided by the mass of solution, multiplied by 100.

Parts per million (ppm)

A unit of concentration equivalent to 1 milligram of solute per liter of water (mg/l) or 1 milligram of solute per kilogram of soil (mg/kg).

Parts per billion (ppb)

A unit of concentration equivalent to 1 gram of solute per 1 billion grams of solution.

Molar concentration calculation

M = moles of solute / volume of solution in liters.

Example of molarity calculation

For a 1 liter solution with 0.5 moles of solute, M = 0.5 mol / 1 L = 0.5 M.

Example of molarity with glucose

A 100.5 mL solution contains 5.10g glucose (C6H12O6).

Moles of Solute

Moles = Molarity × Volume (L)

Conversion of grams to moles

5.10 g glucose x 1 mole glucose / 180.16 g glucose = 0.0283 mol glucose

Conversion of mL to L

100.5 mL = 0.1005 L

Molarity Calculation Example

1 M = 5.10g (glucose) / (180.16 g glucose × 0.1005 L) = 0.282M

Stock Solution

Concentrated solution (i.e., high solute-to-solvent ratio).

Dilution

Preparation of dilute solution (i.e., low solute-to-solvent ratio) by adding solvent to a given volume of stock solution.

Dilution Process

Dilution refers to the process of reducing the concentration of a solution by adding more solvent.

Electrolyte

A solute that produces ions in solution.

Electrolytic Solution

A solution that conducts electricity.

Charge Carriers

Ions

Non-Electrolytes

Substances in which no ionization occurs. There is no conduction of electrical current.

Examples of Non-Electrolytes

Aqueous solutions of sugar, ethanol, ethylene glycol.

Strong Electrolytes

Nearly 100% dissociated into ions and conduct current efficiently.

Examples of Strong Electrolytes

Solutions of NaCl, HNO3, HCl.

Weak Electrolytes

Only partially dissociate into ions and are slightly conductive.

Examples of Weak Electrolytes

Vinegar (aq. solution of acetic acid); tap water.