solving stoichiometry problems

Solving Stoichiometry Problems

Did you know?•Considered individually, the chemical

elements in your body, such as carbon, hydrogen, oxygen, etc., are worth about $20.

•However, when combined into complex molecules, such as fluids, tissues and antibodies, your body is worth over $45 million!

•In the lab, chemists use balances to measure chemicals in grams. Therefore, it is necessary to establish a method for converting a measured mass in grams to a chemical equation, which uses moles as quantities.

•As in mole-mole calculations, the unknown can be either a reactant or a product.

Review• A balanced equation is essential for all calculations

involving amounts of reactants and products.• Problems involving moles and masses can be

solved in basically the same way as mole-mole problems.

• In a mole-mass problem you are asked to calculate the mass of a substance that will react with, or be produced from, a given number of moles of a second substance.

Moles A Moles B Mass B

Tip•Follow these steps to solve mole-mass

problems:

1. Write the balanced equation 2. Determine what is given and what you

are looking for (unknown) 3. Convert moles of the given substance

into moles of the unknown4. Convert moles of the unknown into

grams

Example • Propane is a gas that is commonly used in BBQs. Its formula is C3H8.

Calculate the mass of water that is produced by the combustion of 13.8 mol of propane.

1. Write the balanced equation

2. Determine what is given and what you are looking for (unknown). Unknown: mass of water produced Given: amount of propane = 13.8 mol

3. Convert moles of the given substance into moles of the unknownFrom the equation:

4. Convert moles of the unknown into grams

• So, 2434 g of water can be formed from 13.8 mol of propane.

•In a mass-mole problem, you are asked to calculate the moles of a substance that will react with, or be produced from, a given number of grams of a second substance.

Moles A Moles B Mass B

Tip•Follow these steps to solve mole-mass

problems:1. Write the balanced equation 2. Determine what is given and what you

are looking for (unknown). 3. Convert grams into moles4. Convert moles of the given substance

into moles of the unknown

Example 2 • How many moles of chlorine are required to react completely

with 48.6 g of sodium to produce sodium chloride?1. Write the balanced equation

2. Determine what is given and what you are looking for (unknown).

▫ Unknown: moles of chlorine▫ Given: mass of sodium = 48.6 g

3. Convert grams into moles

4. Convert moles of the given substance into moles of the unknown

• So, 0.416 mol of propane can react with 48.6 g of sodium.

Check Your Understanding•How many moles of potassium chloride

are produced when sufficient chlorine reacts with 78.9 g of potassium iodide?

2KI + Cl2 I2 + 2KCla) 0.26 molb) 0.48 molc) 0.59 mold) 0.98 mol

•Most often, you are given a starting mass and asked to find out the mass of a product. This type of problem is called a mass-mass problem, since you start with a mass and end with a mass. Even though both the given and unknown are in grams, you must convert to moles before using the chemical equation.

Mass A Moles A Moles B Mass B

Tip•Follow these steps to solve mass-mole

problems:1. Write the balanced equation 2. Determine what is given and what you

are looking for (unknown). 3. Convert grams into moles4. Convert moles of the given substance

into moles of the unknown5. Convert moles of the unknown into

grams

Example 3• Calculate how many grams of ammonia can be produced when 206.9

g of nitrogen reacts with hydrogen.1. Write the balanced equation

2. Determine what is given and what you are looking for (unknown). ▫ Unknown: mass of ammonia ▫ Given: mass of nitrogen = 206.9 g

3. Convert grams into moles

4. Convert moles of the given substance into moles of the unknown

5. Convert moles of the unknown into grams

• So, 206.9 g of nitrogen can produce 251.71 g of ammonia.

Check Your Understanding•Carbon monoxide is a toxic gas that can

be produced by incomplete combustion in automobiles.

2CO + O2 2CO2 a) 625 gb) 7.36 g c) 9.85 gd) 16.0 g