chemistry limiting reagents question aqa alevel

Calcium sulfide reacts with calcium sulfate as shown.
CaS + 3 CaSO4 4 CaO + 4 SO2
2.50 g of calcium sulfide are heated with 9.85 g of calcium sulfate until there is no further reaction.
Show that calcium sulfate is the limiting reagent in this reaction.
Calculate the mass, in g, of sulfur dioxide formed.
Mr (CaS) = 72.2
Mr (CaSO4) = 136.2
the mark scheme says this "3 mol of CaSO4 needed for each mol of CaS, and n(CaSO4) is not 3 × n(CaO) (so CaSO4 is the limiting reagent)" which i dont get bc i thought CaS was the limiting reagent bc it's amount is less when we calculate.
Original post by ricecakes1
Calcium sulfide reacts with calcium sulfate as shown.
CaS + 3 CaSO4 4 CaO + 4 SO2
2.50 g of calcium sulfide are heated with 9.85 g of calcium sulfate until there is no further reaction.
Show that calcium sulfate is the limiting reagent in this reaction.
Calculate the mass, in g, of sulfur dioxide formed.
Mr (CaS) = 72.2
Mr (CaSO4) = 136.2
the mark scheme says this "3 mol of CaSO4 needed for each mol of CaS, and n(CaSO4) is not 3 × n(CaO) (so CaSO4 is the limiting reagent)" which i dont get bc i thought CaS was the limiting reagent bc it's amount is less when we calculate.

Have you calculated the moles of CaS and CaSO4 (remember that moles = mass ÷ Mr)?

Now have a look at the molar ratio shown in the equation.
If CaS is the limiting reagent (and CaSO4 in excess), then 3 x (moles of CaS) will be less than the moles of CaSO4.
If CaSO4 is the limiting reagent (and CaS in excess), then (moles of CaSO4) ÷ 3 will be less than the moles of CaS (or alternatively, 3 x (moles of CaS) will be more than the moles of CaSO4).

Once you know which reagent is limiting, use this value of moles in the molar ratio to calculate how much SO2 will be formed
(edited 5 months ago)
Original post by bl0bf1sh
Have you calculated the moles of CaS and CaSO4 (remember that moles = mass ÷ Mr)?

Now have a look at the molar ratio shown in the equation.
If CaS is the limiting reagent (and CaSO4 in excess), then 3 x (moles of CaS) will be less than the moles of CaSO4.
If CaSO4 is the limiting reagent (and CaS in excess), then (moles of CaSO4) ÷ 3 will be less than the moles of CaS (or alternatively, 3 x (moles of CaS) will be more than the moles of CaSO4).

Once you know which reagent is limiting, use this value of moles in the molar ratio to calculate how much SO2 will be formed

THANK YOUU i get it now