5 liters of Butane gas (C4H10) is reacted with plenty of oxygen at 27 oC and 1 atm to produce carbon dioxide and water vapor. Calculate the following:



1- The volume of CO2 produced at 1 atm and 400 oC.

2- The mass of H2O vapor produced.

3- The mol % of the products.

Given:

C4H10 + O2 = CO2 + H2O (Unbalanced equation)

balance the equation.

calculate the moles of butane ( PV=nRT) you started with.

Use the mole ratio in the balanced equation to tell you how many moles of product you get,
then convert (PV=nRT) from moles to volume.

I will be happy to critique your work.

To answer these questions, we need to balance the chemical equation first. The balanced equation for the combustion of butane gas (C4H10) is:

2C4H10 + 13O2 -> 8CO2 + 10H2O

1. To calculate the volume of CO2 produced at 1 atm and 400°C, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

First, we need to calculate the number of moles of CO2 produced. Since we have 5 liters of butane gas, which is C4H10, and the equation is balanced with a 1:8 mole ratio between C4H10 and CO2, we can calculate the number of moles of CO2 produced:

Moles of CO2 = Moles of C4H10 x (8 moles CO2 / 1 mole C4H10)
= 5 x (8 / 1)
= 40 moles

Now, let's convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15
= 400 + 273.15
= 673.15 K

Using the ideal gas law, we can rearrange the equation to solve for V:

V = nRT / P

Plugging in the values:
V = (40 mol) x (0.0821 L·atm/(mol·K)) x (673.15 K) / (1 atm)
V = 224,968 liters (rounded to the nearest whole number).

Therefore, the volume of CO2 produced at 1 atm and 400°C is approximately 225,000 liters.

2. To calculate the mass of H2O vapor produced, we need to use the molar mass of water (H2O). The molar mass of H2O is 18.015 grams/mol.

Using the balanced equation, the mole ratio between C4H10 and H2O is 10:2. So, the number of moles of H2O produced is:

Moles of H2O = Moles of C4H10 x (2 moles H2O / 10 moles C4H10)
= 5 x (2 / 10)
= 1 mole

The mass of H2O produced is given by:
Mass of H2O = Moles of H2O x Molar mass of H2O
= 1 mol x 18.015 g/mol
= 18.015 grams

Therefore, the mass of H2O vapor produced is approximately 18.015 grams.

3. To calculate the mol % of the products, we need to find the moles of each product and then divide by the total moles of all the products.

From the balanced equation, we can see that 1 mole of C4H10 produces 8 moles of CO2 and 10 moles of H2O. So, the total moles of the products are:

Total moles of products = Moles of CO2 + Moles of H2O
= 40 + 10 (as calculated in question 1 and 2)
= 50 moles

The mol % of each product is calculated using the following formula:

Mol % of a product = (Moles of the product / Total moles of products) x 100

For CO2:
Mol % of CO2 = (40 moles / 50 moles) x 100
= 80%

For H2O:
Mol % of H2O = (10 moles / 50 moles) x 100
= 20%

Therefore, the mol % of CO2 is 80% and the mol % of H2O is 20%.