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?
Ca: 40.0 amu F: 19.0 amu H: 1.0 amu S 32.0 amu
CaF2 + H2SO4 --> 2HF + CaSO4
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 =?
How can you write this formula? I'm stuck
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.
To find the maximum number of grams of hydrogen fluoride (HF) that can be obtained, we need to determine the limiting reactant in the given reaction. The limiting reactant is the reactant that will be completely consumed first and determines the maximum amount of product that can be produced.
1. Calculate the molar mass of calcium fluoride (CaF2):
Molar mass of CaF2 = (1 Ca atom x 40.0 g/mol) + (2 F atoms x 19.0 g/mol)
= 40.0 g/mol + 38.0 g/mol
= 78.0 g/mol
2. Calculate the molar mass of sulfuric acid (H2SO4):
Molar mass of H2SO4 = (2 H atoms x 1.0 g/mol) + (1 S atom x 32.0 g/mol) + (4 O atoms x 16.0 g/mol)
= 2.0 g/mol + 32.0 g/mol + 64.0 g/mol
= 98.0 g/mol
3. Determine the number of moles of each reactant:
Number of moles of CaF2 = 11.9 g / 78.0 g/mol
= 0.152 moles
Number of moles of H2SO4 = 10.3 g / 98.0 g/mol
= 0.105 moles
4. Use stoichiometry to determine the moles of HF that can be produced:
From the balanced equation, we can see that 1 mole of CaF2 produces 2 moles of HF.
Since the ratio is 1:2, we can multiply the number of moles of CaF2 by 2 to get the moles of HF.
Moles of HF = 0.152 moles CaF2 x (2 moles HF / 1 mole CaF2)
= 0.304 moles
5. Calculate the mass of HF:
Mass of HF = Moles of HF x Molar mass of HF
= 0.304 moles x (1.0 g/mol + 19.0 g/mol)
= 0.304 moles x 20.0 g/mol
= 6.08 g
Therefore, the maximum number of grams of hydrogen fluoride that can be obtained is 6.08 grams.