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STOICHIOMETRY – Chapter 11

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STOICHIOMETRY – Chapter 11

- Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

- Given : Amount of reactants
- Question:how much of products can be formed.
- Example
- 2 A + 2B 3C
- Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

You will need to use

- molar ratios,
- molar masses,
- balancing and interpreting equations, and
- conversions between grams and moles.
Note: This type of problem is often called "mass-mass."

- Balance the chemical equation correctly
- Using the molar mass of the given substance, convert the mass given to moles.
- Construct a molar proportion (two molar ratios set equal to each other)
- Using the molar mass of the unknown substance, convert the moles just calculated to mass.

A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks)

2 Mg(s) + O2(g) 2 MgO(s)

Mole Ratios:

2 : 1: 2

1) N2 + 3 H2 ---> 2 NH3

Write the mole ratios for N2 to H2 and NH3 to H2.

2) A can of butane lighter fluid contains 1.20 moles of butane (C4H10). Calculate the number of moles of carbon dioxide given off when this butane is burned.

Using the practice question 2) above:

Equation of reaction

2C4H10 + 13O2 8CO2 + 10H2O

Mole ratio

C4H10 CO2

1: 4 [ bases]

1.2 : X [ problem]

By cross-multiplication, X = 4.8 mols of CO2 given off

- Problem 1: 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? [k = 39, Cl = 35.5, O = 16]
2 KClO3 2 KCl + 3 O2

- Use mole ratio
- Get the answer in moles and then
- Convert to Mass. [Simple Arithmetic]
Hello!

If you are given a mass in the problem, you will need to convert this to moles first. Ok?

2 KClO3 2 KCl + 3 O2

2: 3

1.50 : X

X = 2.25mol

Convert to mass

2.25 mol x 32.0 g/mol = 72.0 grams

Cool!

- We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required?
Soln

KClO3 :KCl

2 : 2

X :2.75

X = 2.75mol

In mass: 2.75mol X 122.55 g/mol

= 337 grams zooo zimple!

There are four steps involved in solving these problems:

- Make sure you are working with a properly balanced equation.
- Convert grams of the substance given in the problem to moles.
- Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem.
- Convert moles of the substance just solved for into grams.

Just follow mass-mass problem to the penultimate level

There are four steps involved in solving these problems:

- Make sure you are working with a properly balanced equation.
- Convert grams of the substance given in the problem to moles.
- Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem.
- Convert moles of the substance just solved for into Volume.

No of moles = Volume

Molar volume

Can you remember a similar equation?

The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is: 22.4dm3

Calculate the volume of carbon dioxide formed at STP in ‘dm3' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16)

Solution:

Convert the mass to mole:

Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100gmol-1

Mole = mass/molar mass

3.125/100 = 0.03125mol

As per the equation,

Mole ratio1 : 1

problem0.03125mol X

X = 0.03125mol of CO2

Convert mole to volume [slide 17]

Volume = (0.03125 x 22.4)dm3

=0.7dm3