Chapter 7: Unit 6. Concentration Calculations in Chemical Reactions

Concentrations Calculations in Chemical Reactions

Concentrations and moles calculations in chemical reactions are known as stoichiometry. In most cases, one reactant is used in excess and the other reactant is used in limiting amount. The concentration of the product or the yield of the product is calculated based on the limited reactant amount because it is consumed completely during the chemical reaction.

In the example below, the molarity and the volume of the reactant are used to determine the molarity of the other reactant molarity and/or volume:

Example:

2 Na₃PO₄(aq)    +    3 Ca(NO₃)₂(aq)  à    Ca₃(PO₄)₂(s)     +    6 NaNO₃(aq)

If 355.0 mL of 0.380 M Na₃PO₄ are required to react with 0.135 M Ca(NO₃)₂ to produce Ca₃(PO₄)₂.

Calculate the amount of the volume Ca(NO₃)₂ needed for this reaction.

Solution:

Step 1: number of moles of a given reactant using its molarity and volume:

Number of moles of Na₃PO₄:

355.0 mL are converted into Liters and multiplied with molarity 0.380 mol/L = 0.3550 L X 0.380 mol/L = 0.1349 moles Na₃PO₄

Step 2: using the mole ratio of both reactants from the chemical equation above, one can calculate number of moles of the other reactant:

Mole ratio = 2 moles Na₃PO₄ / 3 moles Ca(NO₃)₂ or 3 moles Ca(NO₃)₂ / 2 moles Na₃PO₄

Moles of Ca(NO₃)₂ = [3 moles Ca(NO₃)₂ / 2 moles Na₃PO₄ ] x [0.1349 moles Na₃PO₄] = 0.20235 moles

Step 3: amount of the volume of Ca(NO₃)₂ can be calculated using its molarity and number of moles:

Amount of Ca(NO₃)₂ needed = [0.20235 moles Ca(NO₃)₂] / [0.135 mol / L Ca(NO₃)₂ = 1.50 L (3 significant figures)

Example:

How many liters of 0.188 M hydrochloric acid HCl would be required to react completely with 2.85 grams of sodium hydroxide NaOH according to the reaction below?

HCl(aq)    + NaOH(aq)  à NaCl(aq)  +  H₂O(l)

Step 1: number of moles of NaOH is calculated with the help of the molar mass of NaOH:

Molar mass of NaOH = Na + O + H = 23.0 + 16.0 + 1.0 = 40.0 g/mol

NaOH number of moles = [(2.85 g) / (40.0 g/mol)] = 0.07125 mol

Step 2: using the mol ratio from the chemical equation:

Mole ratio = 1 mol HCl / 1 mol NaOH = 1:1

HCl moles = NaOH moles = 0.07125 mol

Step 3: with the help of molarity of HCl and its number of moles, its volume can be calculated

Volume of HCl = [0.07125 mol HCl / (0.188 mol/L)] = 0.379 L