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