The equation of simulated photosynthesis is represented by
6CO2(g) + 6H2O(g)-> C6H12O6(s) + 6O2(g)
At 31 degree Celcius, the following equilibrium concentration were found.
[H2O] = 7.91X10^-2M
[CO2] = 9.30X10^-1M
[O2] = 2.40X10^-3M
My question is how to determine the initial mass of CO2 involved in the above reaction?
As I only know how to find the initial concentration of CO2
Why don't you post the entire question in its original form including the numbers?
sry tat i dont know how to insert a picture
To determine the initial mass of CO2 involved in the reaction, we need to use the given concentration and other information.
First, let's calculate the initial moles of CO2 from its initial concentration.
1. Convert the given initial concentration of CO2 to moles:
[CO2] = 9.30 × 10^-1 M
To convert from molarity (M) to moles (mol), we need to multiply by the volume of the solution. However, since the volume is not given, we will assume it is 1 liter for simplicity.
Moles of CO2 = [CO2] × volume
= 9.30 × 10^-1 mol/L × 1 L
= 9.30 × 10^-1 mol
Now we have the initial moles of CO2.
To determine the mass of CO2, we'll use the molar mass of CO2, which is approximately 44 g/mol.
2. Calculate the mass of CO2 using the moles of CO2:
Mass of CO2 = Moles of CO2 × molar mass of CO2
= 9.30 × 10^-1 mol × 44 g/mol
= 40.92 g
Therefore, the initial mass of CO2 involved in the reaction is approximately 40.92 grams.