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
