Question

Propane burns in oxygen to produce carbon dioxide and water. How many grams of oxygen gas are required

to burn 10.0 grams of propane?

Answer 1

To calculate the amount of oxygen needed to burn 10.0 grams of propane, we use the balanced chemical equation for propane's combustion and convert grams to moles to determine that 36.30 grams of oxygen are required.

Step-by-step explanation:

Calculating Oxygen Required for Propane Combustion

The question asks how many grams of oxygen gas are required to burn 10.0 grams of propane. To solve this, we first need to write and balance the chemical equation for the combustion of propane:

C3H8 + 5O2 → 3CO2 + 4H2O

This balanced equation tells us that one mole of propane reacts with five moles of oxygen to produce carbon dioxide and water. Using the molar masses of propane (C3H8) and oxygen (O2), we can calculate the mass of oxygen needed:

1. 
   
3. Calculate the number of moles of propane: (10.0 g) / (44.09 g/mol) = 0.2269 mol
4. 
   
6. Use the stoichiometric ratio from the balanced equation: 0.2269 mol (propane) x (5 mol O2 / 1 mol propane) = 1.1345 mol O2
7. 
   
9. Calculate the grams of oxygen: 1.1345 mol x (32.00 g/mol) = 36.30 grams of O2

Therefore, to burn 10.0 grams of propane completely, 36.30 grams of oxygen are required.

Answer 2

`m_(O_2)=36.3gO_2`

Step-by-step explanation:

Hello!

In this case, since the reaction between propane and oxygen is:

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

Given the 1:5 mole ratio between propane and oxygen, and their molar masses (44.09 g/mol and 32.00 g/mol respectively), we can follow up the shown below stoichiometric setup to compute the required mass of oxygen for the reaction:

`m_(O_2)=10.0gC_3H_8*(1molC_3H_8)/(44.09gC_3H_8)*(5molO_2)/(1molC_3H_8) *(32.00gO_2)/(1molO_2) \\\\m_(O_2)=36.3gO_2`

Best regards!