Processed data table
*Because uncertainty starts with 1, therefore 2 significant numbers is necessary.
Sample Calculations
Calculating the average
(83 + 78 + 78) cm3 3 = 79.66667 cm3
Rounded to two significant figures is 80 cm3
Calculating the uncertainty for my averages
Calculated using residuals
i.e. Largest residual is 83 – 80 = 3 as opposed to 80 – 78 = 2
Absolute uncertainty is + 3
The limiting reagent as when CO2 remains roughly the same (group 3 to 6)
Because Molar Volume of a gas = 24.5 dm3 at SLC (approximately)
NaHCO3 + CH3COOH = CH3COONa + H2O + CO2
1 mole of CO2 is 24500 cm3
and (153 cm3 + 1.0 cm3)* of CO2 is 0.006244 moles because
(153 + 0.6% 24500) 1mol = 0.006285mol + 0.00004
*used the average value out of these groups, uncertainty derived using residuals
Grams of CH3COOH = 0.006285 + 0.00004 mol 60.5
= 0.3802 + 0.003g
n(CH3COOH) =
=
=0.006331 + 0.00004 mol of CH3COOH
Theoretically, 0.006331 + 0.00004 mol of CH3COOH would react with 0.006331 + 0.00004 mol of NaHCO3 , because the ratio is 1:1. But, this may not be the case in our experiment, therefore we need to find the number of moles of NaHCO3
n(NaHCO3) =
= 0.01547 + 0.00005 mol
There are only 0.01547 + 0.00005 mol of NaHCO3 but 0.006331 + 0.00004 mol of CH3COOH is required for the 1.3 g of NaHCO3 to react completely.
CH3COOH is the limiting reagent
We can also say that the number of moles in 10 cm3 of CH3COOH is 0.006331 + 0.00004 mol, as CH3COOH is the limiting reagent and is the one that is completely used up. However, it is noteworthy that the task sheet said that it is a 3-5% acid solution, therefore the uncertainty of the number of moles of CH3COOH could be larger than expected.
Group 1
We know that the number of moles in 10 cm3 of CH3COOH is 0.006331 + 0.00004 mol as seen before. Therefore, we only need to find the number of moles of NaHCO3 to find the limiting reagent in group 1.
n(NaHCO3) =
= 0.003571 + 0.00005 mol
Theoretically, 0.006331 + 0.00004 mol NaHCO3 is required to completely react with 0.006331 + 0.00004 mol of CH3COOH
But there are only 0.003571 + 0.00005 mol of NaHCO3 but 0.006331 + 0.00004 mol of CH3COOH
NaHCO3 is the limiting reagent in group 2