Got my as chemistry practical next week on OCR. Really stuck on it, don't know how to work out how the expected volume of CO2 is greater than that is obtained. Looked at the mark scheme ocr chemistry evalutative task specimen , but its not much help as it only gives the final answers but not any of the working required to get the answers! below is the question.

Procedure
A 50.0 cm3 sample of 0.100 mol dm–3 sulfuric acid was measured using a pipette. This was
added to the flask with the side arm.
0.30 g of magnesium carbonate, MgCO3, was accurately weighed on a digital balance.
The measuring cylinder was filled with exactly 100 cm3 of water and supported upside down in a
trough of water.
The two pieces of apparatus were connected by a delivery tube so that any gas produced in the
side arm flask would be collected in the measuring cylinder. The apparatus was airtight and no
leaks were present.
The rubber bung was removed from the side arm flask. The MgCO3 was quickly added and the
bung replaced.
Bubbles of carbon dioxide were seen collecting in the measuring cylinder.
When no more gas was seen to collect in the measuring cylinder, the volume of gas collected
was recorded as 64 cm3.

Analysis
(a) Construct a fully balanced equation for the reaction between H2SO4 and MgCO3, including
state symbols.

H2SO4 AQ) + MGCO3 (S) --> MGSO4 (AQ) + H20 (L) + CO2 (G) C1 [1]

(b) Calculate the number of moles of H2SO4 and MgCO3 used in the experiment.

moles H2SO4 = 0.005 mol........... C1[1]

(dont know about this one - i got 0.00356 but mark scheme says its 0.0298!)

moles MgCO3 =.......................mol ........... C1[1]

(d) Show by calculation that the expected volume of carbon dioxide is greater than the 64 cm3
obtained.
Assume that one mole of CO2 occupies 24.0 dm3 under the conditions used in the
experiment.

i really don't get how you get 71.4 for the following question:
Show by calculation that the expected volume of carbon dioxide is greater than the 64cm3 obtained. I multiplied 3.56*10-3 the moles of MgCO3 by 24000 and got 85cm3

also I don't understand the mark scheme for the last questions:
d) The students decided that the experiment could be carried out more accurately if both the procedure and measuring apparatus were modified.
i) Draw a labelled diagram to show how the carbonate and acid could be reacted together without needing to remove the bung from the flask.

e) The same method was used to measure the volume of co2 produced when magnesium carbonate reacted with HCL. Suggest with reasons any modifications you need to make if the sulfuric acid used in the procedure was replaced with the same concentration and volume of hydrochloric acid.

To calculate the expected volume of carbon dioxide (CO2) in this experiment, you need to start by finding the number of moles of MgCO3 used. You already have the moles of H2SO4 (sulfuric acid) as 0.005 mol.

First, calculate the moles of MgCO3 using its molar mass:

Molar mass of MgCO3 = (Molar mass of Mg) + (Molar mass of C) + 3*(Molar mass of O)
= 24.31 + 12.01 + 3*(16.00)
= 84.31 g/mol

Given that 0.30 g of MgCO3 was used, you can calculate the moles using the formula:

moles = mass (in grams) / molar mass

moles MgCO3 = 0.30 g / 84.31 g/mol
= 0.00356 mol (rounded to five decimal places)

Now, you can calculate the expected volume of CO2. One mole of any gas occupies 24.0 dm3 at standard temperature and pressure (STP). Therefore, the expected volume can be calculated by multiplying the moles of CO2 by 24.0:

Expected volume of CO2 = moles of CO2 * 24.0 dm3

Expected volume of CO2 = 0.00356 mol * 24.0 dm3/mol
= 0.0854 dm3
= 85.4 cm3 (rounded to three significant figures)

So, the expected volume of carbon dioxide is 85.4 cm3, which is greater than the 64 cm3 obtained in the experiment.

Now, let's move on to the mark scheme for the last questions:

d) To carry out the experiment more accurately, the procedure and measuring apparatus could be modified. Unfortunately, without access to the mark scheme for the question, I cannot provide a detailed explanation of the modifications required. However, the mark scheme should provide a diagram and descriptions of the modifications needed. You can refer to it for a better understanding.

e) If the sulfuric acid used in the procedure is replaced with the same concentration and volume of hydrochloric acid (HCl), some modifications are needed. Hydrochloric acid is a strong acid, and it may react more rapidly and forcefully with the magnesium carbonate. Therefore, you may need to use a different setup or apparatus to control the reaction rate and avoid excessive gas release. Additionally, you may need to consider the stoichiometry and reactivity of HCl compared to H2SO4 and adjust the quantities or concentrations of reagents accordingly. The mark scheme should provide specific suggestions and reasoning for these modifications.