Question

The combustion of methane is a reaction commonly used in chemistry problems due to its ability to fit into multiple topics. So it should not surprise you to see it here as well. How many L of CO2 would be produced if 45 g of CH4 was combusted with ample oxygen in a room that was 90 degrees Celsius and under 1 atm of pressure

Answer 1

Answer:

20.76 L OF CO2 WILL BE PRODUCED BY 45 G OF METHANE.

Explanation:

Equation of the reaction:

CH4 + 02 --------> CO2 + 2H20

Molar mass of methane = ( 12+ 1*4) g/mol = 16 g/mol

Calculate the number of moles present in 45 g of methane

1 mole of methane = 16 g / mol of methane

(45 / 16) mole of methane = 45 g of methane

= 2.8125 moles

Using the ideal gas equation:

PV = nRT

P = 1 atm

n = 2.812 moles

T = 90 C

R = 0.082 L atm/ mol C

V = unknown

So we have:

V = nRT / P

V = 2.8125 * 0.082 * 90 / 1

V = 20.756 L

In the production of CO2 by 45 g of methane, 20.756 L of methane was used.

Then, the volume of CO2 produced by this volume will be 20.756 L since 1 mole of methane produces 1 mole of CO2.

In other words;

1 mole of CH4 = 1 mole of CO2

22.4 dm3 of CH4 = 22.4 dm3 of CO2

20.76 DM3 = 20.76 dm3

The volume of CO2 produced will therefore be 20.76 L