Aim: To investigate the reaction between sodium bicarbonate powder and ethanoic acid and discover the concept of limiting reagents.

*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 would 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

Group 2

We know that the number of moles in 10 cm3 of CH3COOH is 0.006331 mol as seen before. Therefore, we only need to find the number of moles of NaHCO3 to find the limiting reagent in group 2.

n(NaHCO3)        =

                = 0.005952 + 0.00005 mol

Theoretically, 0.006331 + 0.00004 mol NaHCO3 is required to completely react with 0.006331 + 0.00004 mol of CH3COOH

There are only 0.005952 + 0.00005 mol of NaHCO3 but 0.006331 + 0.00004 mol of CH3COOH

 NaHCO3 is the limiting reagent in group 2

Now we know that

We want to know how much sodium bicarbonate in grams is required to react perfectly with 10 cm3 of diluted CH3COOH which contains 0.006331 mol CH3COOH

 For it to react ideally, 0.006331 mol of NaHCO3 has to react with 0.006331 mol of CH3COOH

NaHCO3) = 0.006331 mol

 84

                = 0.5318 grams