If 4 grams of Hydrogen gas (H2) reacts with 76 grams of fluorine gas (F2), how many grams of hydrofluoric acid (HF) will be produced? If you are unsure how to work out this problem, see a similar example linked HERE. a 80 grams of HF will be produced. b 304 grams of HF will be produced. c 72 grams of HF will be produced. d 19 grams of HF will be produced.
b 304 grams of HF will be produced.
The balanced chemical equation for the reaction is:
H2 + F2 -> 2HF
First, we need to determine the limiting reactant by calculating the moles of each reactant:
- Moles of H2 = 4 g / 2 g/mol = 2 mol
- Moles of F2 = 76 g / 38 g/mol = 2 mol
Since the moles of both reactants are equal, F2 is the limiting reactant as it has a higher molar mass. Therefore, all the F2 will react with H2 to produce HF.
Now, we need to calculate the mass of HF produced using the stoichiometry of the reaction:
- Moles of HF = 2 mol F2 x (2 mol HF / 1 mol F2) = 4 mol HF
- Mass of HF = 4 mol HF x 20 g/mol = 80 g
Therefore, 80 grams of HF will be produced.