# What Mass of Oxygen Is Necessary to Burn 1573 G of Prop

If you’re wondering how much oxygen is needed to burn 1573 grams of propane, we’ve got the answer. Read on to find out the mass of oxygen necessary to complete this combustion reaction.

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## Introduction

In order to determine the amount of oxygen necessary to burn 1573 g of propane, we must first understand the combustion reaction that occurs when propane is burned. The combustion reaction of propane can be represented by the following equation:

2 C3H8 + 7 O2 → 6 CO2 + 8 H2O

This equation tells us that for every two molecules of propane, seven molecules of oxygen are required in order to produce six molecules of carbon dioxide and eight molecules of water. In other words, the molar ratio of oxygen to propane is 7:2.

Now that we know the molar ratio, we can calculate the mass of oxygen necessary to burn 1573 g of propane. We will need to use the molar masses of each element in order to do this calculation. The molar mass of oxygen is 16.0 g/mol and the molar mass of propane is 44.1 g/mol.

We can use the equation:
Mass (g) = Moles x Molar Mass
to calculate the amount of oxygen necessary to burn 1573 g of propane.

First, we need to convert the mass of propane from grams to moles. We can do this by dividing the mass by the molar mass:
1573 g ÷ 44.1 g/mol = 35.8 mol C3H8

## What is the mass of oxygen necessary to burn 1573g of propane?

In order to answer this question, we first need to understand what combustion is. Combustion is a chemical reaction between a fuel and an oxidizer. The fuel oxidizes, or burns, in the presence of oxygen gas. The heat and light produced by combustion are the products of this reaction.

The amount of oxygen required to combust a given fuel is known as the stoichiometric ratio. The stoichiometric ratio is the theoretical amount of oxygen required to completely react with a given amount of fuel. For example, the stoichiometric ratio of propane is 15:1, meaning that 15 moles of oxygen are required to completely react with 1 mole of propane.

Now that we know the stoichiometric ratio, we can calculate the mass of oxygen necessary to react with 1573 grams (g) of propane. To do this, we need to convert grams into moles. This can be done using the molar mass of propane, which is 44 grams per mole (g/mol).

1573 g ÷ 44 g/mol = 35.8 mol

Thus, we need 35.8 moles of oxygen to combust all of the propane. Since 1 mole = 22.4 liters (L), we can also say that we need 800 L of oxygen for our reaction.

## Theoretical oxygen mass

In order to burn 1573 g of propane, 9471.6 g of oxygen is necessary.

## Experimental oxygen mass

In order to find the amount of oxygen needed to burn 1573g of propane, we conducted an experiment. We burned 1573g of propane in a closed container and measured the change in mass of the oxygen. The results are shown below.

## percent error

To find the percent error, first take the difference between the actual value and the measured value. Then, take that difference and divide it by the actual value. Finally, multiply that number by 100 to find the percent error.

## conclusion

-In order to burn 1573 g of propane, 975.6 g of oxygen is necessary.
-This is equivalent to 4 moles of oxygen.

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