Sulfuric acid, H2SO4, reacts with CaF2 according to:

CaF2 + H2SO4 ---> CaSO4 + 2 HF

What is the minimum amount, in g, of impure H2SO4 (which is 83.5% H2SO4 by mass) that would be required to produce 16.2 g of HF when reacted with excess CaF2 ?

Could someone walk me through the steps?

calculate the moles of HF from the mass.

calculate the moles of sulfuric acid needed (1/2 the moles of HF).
calculate the mass of sulfuric acid from the moles of sulfuric acid.
divide by .835

To determine the minimum amount of impure H2SO4 needed to produce 16.2 g of HF when reacted with excess CaF2, we need to use stoichiometry calculations. Here's how you can go about solving the problem step by step:

Step 1: Write down the balanced chemical equation.
CaF2 + H2SO4 ---> CaSO4 + 2 HF

Step 2: Calculate the molar mass of HF.
The molar mass of HF is 1.01 g/mol for hydrogen (H) and 19.0 g/mol for fluorine (F). Adding these together, the molar mass of HF is 20.01 g/mol.

Step 3: Convert the given mass of HF to moles.
Divide the given mass of HF (16.2 g) by its molar mass to convert it to moles:
16.2 g HF / 20.01 g/mol = 0.8096 mol HF

Step 4: Use the stoichiometry of the balanced equation to find the moles of H2SO4 needed.
From the balanced equation, we see that 1 mole of CaF2 reacts with 1 mole of H2SO4 to produce 2 moles of HF. Therefore, the moles of H2SO4 needed would be half of the moles of HF produced.
0.8096 mol HF x (1 mol H2SO4 / 2 mol HF) = 0.4048 mol H2SO4

Step 5: Convert the moles of H2SO4 to grams.
To do this, we need to know the molar mass of H2SO4, which is 98.09 g/mol. Multiply the moles of H2SO4 by its molar mass to get the mass in grams:
0.4048 mol H2SO4 x 98.09 g/mol = 39.7 g

However, keep in mind that the H2SO4 used is impure and contains 83.5% H2SO4 by mass. So now we need to adjust for this impurity:

Step 6: Calculate the mass of pure H2SO4.
Multiply the mass of impure H2SO4 (39.7 g) by its purity percentage (83.5% converted to decimal):
39.7 g x 0.835 = 33.15 g

Therefore, the minimum amount of impure H2SO4 required to produce 16.2 g of HF when reacted with excess CaF2 is approximately 33.15 grams.