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# 12.2: Stoichiometric Calculations

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Based on the balanced chemical equation, be able to calculate the masses of reactants or products generated in a given reaction.
• Based on the balanced chemical equation, be able to calculate the moles of reactants or products generated in a given reaction.
• Understand how to convert between masses and moles in a chemical reaction using mole ratios and molar masses.

### Recalling Prior Knowledge

1. How much hydrogen is needed to form 3.1 moles of tin according to the following reaction?
SnO2 + 2 H2 → Sn + 2H2O

## Mole Ratios, Molar Masses, and Chemical Equations

How can we measure out a known amount of a reactant, since actually counting atoms and molecules is not a practical approach? How can we tell what amount of product was generated in a reaction? In most cases, the mass of a reactant or product is a relatively easy quantity to measure. Recall that the molar mass of a given chemical species can be determined by referencing the periodic table. If we know the identity of the substance we wish to measure, molar mass can be used as a conversion factor between mass and amount (in moles). For any given chemical reaction, we can describe the following relationships in the Figure below.

This image depicts how moles, mass, and mole ratios are related for a given chemical equation.

Example 12.3

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

How many grams of each reactant are needed to produce 0.500 mol of silver chloride?

First, we need to relate the mass of silver nitrate to the amount in moles of the product silver chloride. This can be accomplished by using a series of conversion factors in which all units cancel except for those of the desired answer (grams of silver nitrate). To do this, we will need the mole ratio between these two reaction components and the molar mass of silver nitrate. Then, we can perform the following calculation:

$\mathrm{0.500 \: mol \: AgCl \times (\frac{1 \: mol \: AgNO_3}{1 \: mol \: AgCl})(\frac{169.87 \: g \: AgNO_3}{1 \: mol \: AgNO_3})=84.9 \: g \: AgNO_3}$

In order to produce 0.500 moles of AgCl, we would need to start with 84.9 g of silver nitrate. A similar calculation can be performed to determine the necessary mass of sodium chloride.

$\mathrm{0.500 \: mol \: AgCl \times (\frac{1 \: mol \: NaCl}{1 \: mol \: AgCl})(\frac{58.44\:g\:NaCl}{1 \: mol \: NaCl})=29.2\:g\:NaCl}$

Mass Reactants$\mathrm{\leftrightarrow}$Moles Reactants$\mathrm{\leftrightarrow}$Moles Products$\mathrm{\leftrightarrow}$Mass Products

In the chemistry lab, we frequently need to calculate the relationship between two reactants or products in a chemical reaction. For example, we may know the mass of one reactant and want to know how much of a given product will be generated if the reactant is fully consumed. We may also wish to know how much of a second reactant is required to fully react with the first reactant. These types of questions can be answered by using molar masses and mole ratios as conversion factors. We will illustrate this process with an example.

Example 12.4

How many grams of lead(II) chloride would be produced if 1.67 g of lead(II) nitrate is allowed to react completely in the presence of a sodium chloride solution? How many grams of sodium chloride would be consumed in the process?

Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)

First, we need to relate grams of lead(II) chloride to grams of lead(II) nitrate. We can set up the following expression, using the molar masses of each component and their mole ratio, obtained from the balanced equation:

1.67 Pb(NO3)2 × ($\mathrm{\frac{1 \ mol \ Pb(NO_3)_2}{331.2 \ g \ Pb(NO_3)_2}}$)($\mathrm{\frac{1 \ mol \ PbCl_2}{1 \ mol \ Pb(NO_3)_2}}$)($\mathrm{\frac{278.11 \ g \ PbCl_2}{1 \ mol \ PbCl_2}}$)=1.40 g PbCl2

Therefore, 1.40 g of lead(II) chloride would be produced if 1.67 g of lead(II) nitrate is fully consumed. The amount of NaCl that would be used in this process can be calculated as follows:

1.67 Pb(NO3)2 × ($\mathrm{\frac{1 \ mol \ Pb(NO_3)_2}{331.2 \ g \ Pb(NO_3)_2}}$)($\mathrm{\frac{2 \ mol \ NaCl}{1 \ mol \ Pb(NO_3)_2}}$)($\mathrm{\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl}}$) = 0.589 g NaCl

In order to fully consume 1.67 g of lead(II) nitrate, we would need at least 0.589 g of NaCl.

## Lesson Summary

• Using molar masses and mole ratios, we can find the relationships between the masses of various reaction components for a given reaction.

## Lesson Review Questions

1. Aluminum reacts with oxygen to produce aluminum oxide according to the following equation: 4Al + 3O2 → 2Al2O3
1. How many grams of O2 are needed to produce 5 moles of Al2O3?
2. How many grams of Al2O3 are produced from the reaction of 5 moles of Al?
3. How many grams of Al are needed to produce 86.0 grams of Al2O3?
2. How many grams of each reactant are needed to produce 0.500 mol of barium sulfate according the following equation? BaCl2 + Na2SO4 → BaSO4 + 2NaCl
3. How many grams of each reactant are needed to produce 28.6 grams copper (II) sulfide by the following reaction? Cu + SO2 → CuS + O2

1. Stoichiometry Calculator: http://mmsphyschem.com/stoichiometry.htm
2. Practice Balancing Chemical Equations:
3. Chemical equation balances: http://www.personal.psu.edu/jzl157/balance.htm

## Points to Consider

• Our study of masses and amounts, as described by a given chemical equation, has assumed that mass is conserved for all chemical processes. How might you determine experimentally that this is the case for a given chemical reaction?
• So far, we have assumed that all of the reactants are utilized in the formation of products during each reaction. Can you think of a case in which one or more reactants would not be completely consumed?

## Date Created:

Sep 09, 2013

May 06, 2015
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