Phosphoric acid can be prepared by reaction of sulfuric acid with "phosphate rock" according to the equation: ca3(po4)2 + 3h2so4 → 3caso4 + 2h3po4 suppose the reaction is carried out starting with 125 g of ca3(po4)2 and 75.0 g of h2so4. which substance is the limiting reactant

The answer and explanation are as follows
According to the equation Ca3(PO4)2 reacts with H2SO4 in the mole ration of 1 is to 3 (1:3)...
You should work in mole hence have to convert the given mass to mole by dividing by the Mr value if each compound...(Mr obtained in periodic table..).
Mr of Ca3(PO4)2 = 310
Mr of H2SO4 = 98.1

Therefore
mole of Ca3(PO4)2 = (125÷310) = 0.403

And mole of H2SO4 = (75÷98.1) = 0.764

Using mole ratio;
1 mole of Ca3(PO4)2 reacts with 3 mole of H2SO4.

So 0.403 mole of Ca3(PO4)2 reacts with (3×0.403) mole of H2SO4...
= 1.209 mole

However note that there is only 0.764 mole of H2SO4 in the solution so it is the limiting reagent ...

Note that it will be the 0.764 mole of acid that will react with (1÷3)×0.764 of the posphate compound...

Hope that it is clear

The answer to that would be magnesium