Hydrofluoric acid (HF) can be prepared according to the following equation: CaF2 + H2SO4 2 HF + CaSO4.
How many grams of HF can be prepared from 75.0 g of H2SO4 and 63.0 g of CaF2?
In the above problem, which reactant is limiting and which is in excess?

Given the balanced equation:

( Reaction type : double replacement)

CaF2 + H2SO4 → CaSO4 + 2HFI

We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.

This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.

For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.

For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.

Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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