A mixture containing 11.9 g of calcium fluoride and 10.3 g of sulfuric acid is heated to drive off hydrogen fluorid. Given the following reaction what is the me maximum number of grams of hydrogen fluoride that can be obtied?

Question

Ca: 40.0 amu F: 19.0 amu H: 1.0 amu S 32.0 amu
CaF2 + H2SO4 --> 2HF + CaSO4

Answer 1

This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. I do these the long way.

CaF2 + H2SO4 --> 2HF + CaSO4

mols CaF2 = grams/molar mass = ?
mols H2SO4 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols CaF2 to mols HF
Do the same and convert mols H2SO4 to mols HF.
It is quite likely that these two values will not agree; the correct value in LR problems is ALWAYS the smaller number and the reagent responsible for that number is the LR.

Use the smaller number of mols HF, then
g HF = mols HF x molar mass HF =?

Answer 2

How can you write this formula? I'm stuck

Answer 3

Just follow each step as I wrote them. It will come out. It a horrendous "formula" if you try to write it all at one time. Show what you have done? Where are you stuck? What do you not understand about the next step.