# How many moles of H2O formed when 5.00 moles of C3H8 completely reacts?

To determine the number of moles of H2O formed when 5.00 moles of C3H8 completely reacts, we first need to write the balanced chemical equation for the complete combustion of propane (C3H8). The balanced equation is:

C3H8 + 5O2 → 3CO2 + 4H2O

This equation shows that for every 1 mole of C3H8 that reacts, 4 moles of H2O are produced. Since we are starting with 5.00 moles of C3H8, we can calculate the moles of H2O formed as follows:

Moles of C3H8: 5.00 moles

Moles of H2O per 1 mole C3H8: 4 moles

Moles of H2O formed = Moles of C3H8 x Moles of H2O per 1 mole C3H8

= 5.00 moles x 4 moles/1 mole

= 20.0 moles

Therefore, when 5.00 moles of C3H8 completely reacts, 20.0 moles of H2O are formed.

## Detailed Explanation

Combustion reactions involve the reaction of a hydrocarbon fuel with oxygen gas to produce carbon dioxide and water. For hydrocarbons containing carbon, hydrogen and oxygen, the general combustion reaction is:

CxHy + zO2 → xCO2 + y/2 H2O

Where x, y and z are coefficients that can be determined by balancing the chemical equation. For propane (C3H8), the balanced combustion reaction is:

C3H8 + 5O2 → 3CO2 + 4H2O

This balanced equation shows us that for every 1 mole of propane (C3H8) that reacts:

• 3 moles of CO2 are produced
• 4 moles of H2O are produced
• 5 moles of O2 are consumed

So if we start with 5.00 moles of C3H8, and it fully reacts according to the balanced equation, we can calculate the moles of H2O formed as:

Moles of C3H8: 5.00 moles

Moles of H2O per 1 mole C3H8: 4 moles (from balanced equation)

Moles of H2O formed = Moles of C3H8 x Moles of H2O per 1 mole C3H8

= 5.00 moles x 4 moles/1 mole
= 20.0 moles

Therefore, starting with 5.00 moles of propane (C3H8) and allowing it to fully react, 20.0 moles of water (H2O) will be formed.

## Stoichiometric Calculations

The above calculation is an example of a stoichiometric calculation. Stoichiometry involves using molar ratios from a balanced chemical equation to calculate the amounts of products and reactants.

Some key points about stoichiometric calculations:

• A balanced chemical equation provides the molar ratio between the reactants and products
• The coefficients in a balanced equation represent the molar ratios
• We can use these molar ratios to convert between amounts of reactants and products
• For the example reaction:
• The ratio of C3H8:H2O is 1:4
• So if we know the moles of C3H8, we can calculate the moles of H2O formed
• The same method can be used to calculate moles of any product or reactant species

Stoichiometric calculations allow us to quantitatively relate the amounts of reactants and products involved in a chemical reaction.

### Sample Stoichiometric Calculation

Let’s look at one more example stoichiometric calculation:

Methane (CH4) reacts with oxygen to produce carbon dioxide and water. The balanced equation is:

CH4 + 2O2 → CO2 + 2H2O

If 0.500 moles of methane fully reacts, how many moles of carbon dioxide are produced?

Using the molar ratio from the balanced equation:

• 1 mole CH4 produces 1 mole CO2

Moles of CH4 given: 0.500 moles

Moles of CO2 per 1 mole CH4: 1 mole (from balanced equation)

Moles of CO2 produced = Moles of CH4 x Moles of CO2 per 1 mole CH4

= 0.500 moles x 1 mole/1 mole
= 0.500 moles

Therefore, if 0.500 moles of methane reacts completely, 0.500 moles of CO2 will be produced.

## Limiting Reactants

Up until now, we have assumed that the reaction goes to completion and the reactant fully reacts. However, in reality, one reactant often limits how much product can form.

This reactant is called the limiting reactant, because it limits the maximum amounts of products generated.

To identify the limiting reactant:

1. Determine the moles of each reactant provided
2. Use stoichiometry and the mole ratios to calculate the maximum moles of product possible from each reactant
3. The reactant generating the smallest amount of product is the limiting reactant

Once the limiting reactant is consumed, the reaction stops, even if there are moles of other reactants remaining.

### Example with Limiting Reactant

Using the methane combustion reaction again:

CH4 + 2O2 → CO2 + 2H2O

If 2.00 moles of CH4 and 4.00 moles of O2 are present, which is the limiting reactant?

Moles of CH4 provided: 2.00 moles

From ratio in equation:
– 1 mole CH4 consumes 2 moles O2
– So 2.00 moles CH4 will consume 2.00 * 2 = 4.00 moles O2

Moles of O2 provided: 4.00 moles

From ratio in equation:
– 2 moles O2 reacts with 1 mole CH4
– So 4.00 moles O2 will react with 4.00/2 = 2.00 moles CH4

Oxygen will be completely consumed after reacting with 2.00 moles CH4.
But we only have 2.00 moles CH4 present.
Therefore, CH4 is the limiting reactant.

Once the 2.00 moles of CH4 fully reacts, the reaction will stop, leaving 2.00 moles of unused O2.

## Conclusion

In summary:

• Combustion of propane follows the reaction:
C3H8 + 5O2 → 3CO2 + 4H2O
• Using the molar ratios from the balanced equation, we can calculate the moles of any product formed, based on the given moles of reactant
• This is an example of a stoichiometric calculation
• If multiple reactants are present, the limiting reactant is the reactant that generates the least amount of product
• The reaction stops once the limiting reactant is fully consumed

For the original example with 5.00 moles of C3H8:

• No limiting reactant was specified, so we assume the reaction goes to completion
• Using stoichiometry, 20.0 moles of H2O will be formed

Stoichiometric calculations allow us to quantitatively relate reactants and products using chemical equations. This is fundamental to many important chemistry calculations.