A level question help

Magnesium carbonate is a common antacid that is used to react with and neutralise excess stomach acid, HCl, with the following equation:
MgCO3 (s) + 2HCl (aq) → MgC|2(aq) +
CO2 (g) + H2O (1)
The total mass of the reactants is 9.91 g.
What volume ofCO2 produced at RTP?
Can someone please tell me steep by step what the process is

Are you sure that is the entire question? Is there previous parts to this question that give more information

It’s a question on save my exam and that’s the only info they give

can you send me the link to it

Magnesium carbonate is a common antacid that is used to react with and neutralise excess stomach acid, HCl, with the following equation:
MgCO3 (s) + 2HCl (aq) → MgC|2(aq) +
CO2 (g) + H2O (1)
The total mass of the reactants is 9.91 g.
What volume ofCO2 produced at RTP?
Can someone please tell me steep by step what the process is

Magnesium carbonate is a common antacid that is used to react with and neutralise excess stomach acid, HCl, with the following equation:
MgCO3 (s) + 2HCl (aq) → MgC|2(aq) +
CO2 (g) + H2O (1)
The total mass of the reactants is 9.91 g.
What volume ofCO2 produced at RTP?
Can someone please tell me steep by step what the process is

As above, you need more information to get an answer as it isn’t clear if a reagent is in excess.
Save my exams does often take past paper questions to use in their sheets, but I’d be unsurprised if they only took part of the question and thus failed to give you all the stuff you need.

Oh I got it. This is kind of a weird question but it is possible
Based on the equation, we know that the mole ratio of the reactants is 1:2 (so for every one mole of MgCO3 you have you have 2 moles of HCl reacting.
We also know that the mass of each reactant is equal to the moles x the Mr (M=nxMr)
So the total mass of the reactants is equal to the sum of the mass of the 2 reactants. since we have a mole ratio we can just let the number of moles be x which gives us 9.91=(x*84.3)+(2x*36.5).
(84.3 is the Mr of MgCO3, 36.5 is the Mr of HCl and I used 2x as we have 2 times the moles of HCl)
If we solve for x, then we get x as 991/15730 (the number of moles of MgCO3)
Again, using the mole ratio, we see that for every mole of MgCO3 we have, We have 1 mole of CO2. So the number moles of CO2 is also 991/15730.
Now to work out the volume of CO2, we need to use the fact that at RTP, 1 mol of a gas occupies 24dm^3. So if we have 991/15730 moles of CO2, then we just mutlipy this by 24 to get 1.51dm^3 of CO2.
Hopefully that's right. If it's wrong or it doesn't make sense at any point, just let me know

Oh I got it. This is kind of a weird question but it is possible
Based on the equation, we know that the mole ratio of the reactants is 1:2 (so for every one mole of MgCO3 you have you have 2 moles of HCl reacting.
We also know that the mass of each reactant is equal to the moles x the Mr (M=nxMr)
So the total mass of the reactants is equal to the sum of the mass of the 2 reactants. since we have a mole ratio we can just let the number of moles be x which gives us 9.91=(x*84.3)+(2x*36.5).
(84.3 is the Mr of MgCO3, 36.5 is the Mr of HCl and I used 2x as we have 2 times the moles of HCl)
If we solve for x, then we get x as 991/15730 (the number of moles of MgCO3)
Again, using the mole ratio, we see that for every mole of MgCO3 we have, We have 1 mole of CO2. So the number moles of CO2 is also 991/15730.
Now to work out the volume of CO2, we need to use the fact that at RTP, 1 mol of a gas occupies 24dm^3. So if we have 991/15730 moles of CO2, then we just mutlipy this by 24 to get 1.51dm^3 of CO2.
Hopefully that's right. If it's wrong or it doesn't make sense at any point, just let me know

Whilst your calculations are fine, the answer is only correct under the assumption that the mole ratio is perfectly stoichiometric and that when it specifies the reactants, it refers to the HCl solute and not the total mass of HCl solution used (which, mind you would be a very small amount indeed if it is the latter).
More information really is needed imo.