Stoichiometric calculations in Enem

O calculation stoichiometric is a very recurrent theme in all editions of Enem and is directly or indirectly present in several other contents of Chemistry, such as:

Solutions
thermochemistry
Chemical Kinetics
Chemical balance
Electrochemistry
Radioactivity
Study of gases
Organic functions

In this text you will have access to very important tips to solve simple stoichiometric calculations in Enem:

1st Tip: Fundamental knowledge to develop stoichiometric calculation

Lavoisier's Law: the sum of the masses of the reactants is equal to the sum of the masses of the products.

A + B → C + D

mA + mB = mC + mD

Proust's Law: The mass proportion of each of the participants in the reaction is always the same.

A + B → C + D

bad + MB = mC + mD
mA' mB' mC' mD'

Mol (amount of matter): according to Avogadro, in one mole, we always have 6.02.10²³ entities (molecules, atoms, ions etc.).

1mol6.02. 10²³

Calculation of molar mass: the molar mass, calculated by the substance formula (XaYb), is the sum of the results of multiplying the quantity of each chemical element by its atomic mass.

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Molar mass = a.mass of X (in the Periodic Table) + b.mass of Y (in the Periodic Table)

Molar mass: equivalent to the mass in grams corresponding to 6.02.10²³ substance entities.

1mol6.02. 10²³mass in grams (molar)

Molar volume: equivalent to 22.4 liters, which refer to the space occupied by 6.02.10²³ substance entities:

1mol6.02. 10²³mass in grams (molar) 22.4L

Balancing chemical equations: coefficients that make the number of atoms of all chemical elements equal in reactants and products.

2nd Tip: Fundamental steps to solve a stoichiometric calculation

Remove the data provided by the exercise;
Write the chemical equation if the exercise did not provide it;
Balance the equation;
The coefficients used in the balancing must be used to know the stoichiometric proportions between the participants;
Set up rules of three that relate the information present in the statement, elements of the equation and its balance.

3rd Tip: Fundamental relationships in stoichiometric calculation

In every rule of three that is assembled in a stoichiometric calculation exercise, we can make the following relationships

Volume————————-mol

Volume————————--Volume

Mass—————————mols

Mass————————— Mass

Mass—————————No. of entities

mol—————————No. of entities

Volume—————————No. of entities

Volume—————————mass

Tip 4: How to proceed in an exercise involving successive reactions

Successive reactions are reaction steps that form a single reaction. When they are part of the exercise, we must, before performing the stoichiometric calculation, form a single reaction.

For this, we must cancel the substance that appears in the reagent of one and in the product of the other. For example:

S + O₂ → OS₂

ONLY₂ + O₂ → OS₃

ONLY₃ + H₂O → H₂ONLY₄

canceling the OS₂ and the OS₃, we have the following reaction:

S + 3/2O₂ + H₂O → H₂ONLY₄

5th Tip: How to proceed in an exercise that involves a reagent in excess and limiting

We know that an exercise involves excess and limiting whenever in the statement we have the presence of the mass of the two substances that make up the reactants. To develop stoichiometric calculations, we must always use the bound mass.

To find out the mass of the limiting reactant, just divide the molar mass of each substance, multiplied by its stoichiometric coefficient in the equation, and divided by the mass given by the exercise.

For example, if we have a chemical reaction of 50 grams of NaCl with 50 grams of CaBr₂:

2 NaCl + 1 CaBr2 → 2 NaBr + 1 CaCl₂

2.58,5 = 1. 200
50 50

2,34 = 4

The largest value of this division always corresponds to the excess reagent, while the smallest value always corresponds to the limiting reagent.

6th Tip: How to proceed in an exercise involving purity

Stoichiometric calculation exercises that involve purity or impurity have in the statement the percentage referring to the pure or impure part of a sample. So, first of all, we must calculate what is the really pure mass of the sample, as it alone gives rise to the product of a reaction.

For example, if we have 70 grams of a sample and 20% of it is impure, then 80% of it is pure. So, we set up a rule of three to determine the mass in grams that is pure:

70g100%

xg80%

100.x = 70.80

100x = 5600

x = 5600
100

x = 56 grams of pure dough.

7th Tip: How to proceed in an exercise that involves Yield

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Yield is related to the actual amount, in grams, of a product that has been formed from a certain mass of reactant. The exercise usually tells us how much mass has been formed. We must then calculate the mass of the product with the mass of the supplied reagent and play the rule of three below:

Calculated product mass 100%

Product mass x%
provided by
exercise

For example, in the reaction of 40 grams of carbon with oxygen, 15 grams of carbon dioxide were formed. What will the reaction yield?

1 C + 1 O₂→ 1 CO₂

1.12g of carbon 1.44g of CO₂
40 g of carbonx

12.x = 40.44
12x = 1760
x = 1760
12
x = 146.6 g of CO₂

Then we determine the yield:

146.6 g100%
15gx%

146.6x = 1500
x = 1500
146,6
x= 10.2%

Follow the resolution now of two examples:

Example 1: (Enem) Currently, polluting emission purification systems are being required by law in an increasing number of countries. Controlling gaseous sulfur dioxide emissions from burning coal, which contains sulfur, can be made by the reaction of this gas with a suspension of calcium hydroxide in water, forming a non-polluting product of the air. The burning of sulfur and the reaction of sulfur dioxide with calcium hydroxide, as well as the masses of some of the substances involved in these reactions, can be represented as follows:

sulfur (32 g) + oxygen (32 g) → sulfur dioxide (64 g)

sulfur dioxide (64 g) + calcium hydroxide (74 g) → non-polluting product

In this way, to absorb all the sulfur dioxide produced by burning a ton of coal (containing 1% sulfur), it is sufficient to use a calcium hydroxide mass of about:

a) 23 kg.

b) 43 kg.

c) 64 kg.

d) 74 kg.

e) 138 kg.

Resolution:

Data provided by the exercise:

1 ton of coal (C)
In coal, we have 1% sulfur (purity)
What is the mass of calcium hydroxide?

1^O Step: Build an equation only from the successive reactions provided:

S + O₂ → OS₂

ONLY₂ + Ca(OH)₂ → CaCO₃+ H₂s

Cutting what is repeated, we have the following reaction:

S + 1/2O₂+ Ca(OH)₂ → CaCO₃+H₂s

NOTE: This step can be neglected, as the exercise involves only sulfur and calcium hydroxide

2^O Step: Calculate the mass of sulfur present in 1 ton of coal, remembering that 1% is sulfur, then:

1t of 100% coal
x sulfur1%

100x = 1
x = 1
100
x = 0.01 t or 10 kg of sulfur

3^O Step: From the sulfur mass, we can calculate the calcium hydroxide mass. In this stoichiometric calculation, we will only list masses:

S Ca(OH)₂
1.32g 1.74g
10 kg

32.x = 74.10
x = 740
32
x = 23.125 kg of butane gas

Example 2: (Enem) In Japan, a national movement to promote the fight against global warming carries the slogan: 1 person, 1 day, 1 kg of CO₂ love us! The idea is for each person to reduce the amount of CO by 1 kg₂ issued every day, through small ecological gestures, such as reducing the burning of cooking gas. An ecological hamburger? And for now! Available in: http://lqes.iqm.unicamp.br. Accessed on: Feb 24 2012 (adapted).

Considering a complete combustion process of a cooking gas composed exclusively of butane (C₄H₁₀), the minimum amount of this gas that a Japanese must stop burning to meet the daily goal, just with this gesture, is it?

Data: CO₂ (44 g/mol); Ç₄H₁₀ (58 g/mol).

a) 0.25 kg.

b) 0.33 kg.

c) 1.0 kg.

d) 1.3 kg.

e) 3.0 kg.

Resolution:

The data provided by the exercise were: