chemistry help....?

Question

A mixture consisting of 12.2 g of calcium fluoride, CaF2, and 13.2 g of sulfuric acid, H2SO4, is heated to drive off hydrogen fluoride, HF.
CaF2(s) + H2SO4(l) ==> 2HF(g) + CaSO4(s)

What is the maximum number of grams of hydrogen fluoride that can be obtained?

Answer 1

CaF2(s) + H2SO4(l) ==> 2HF(g) + CaSO4(s)

Step 1: Find the # of moles that you have of each reactant.
CaF2 has a molar mass of 78g/mol (40 for Ca + 19 * 2 for F)
H2SO4 has a molar mass of 98 g/mol (4 * 16 for O, 2 * 1 for H, 32 for S).
Thus, you have 12.2 / 78 = 0.1564 moles of Ca2F and 13.2 / 98 = 0.1347 moles of H2SO4

Therefore, H2SO4 is the limiting reactant.

Step 2: Determine how many moles of HF were created.
For each mole of H2SO4 going in, 2 moles of HF come out. 0.1347 moles of H2SO4 * 2 moles = .2694 moles HF

Step 3: Determine the molar mass of HF, then find the # of grams.
HF's molar mass is 20 (1 for H, 19 for F). .2694 mol * 20 g/mol = 5.388 g HF. (solution!)