Understanding the Reactions
* Hydrogen + Fluorine → Hydrogen Fluoride (HF)
* The reaction is: H₂ + F₂ → 2HF
* Hydrogen + Bromine → Hydrogen Bromide (HBr)
* The reaction is: H₂ + Br₂ → 2HBr
Key Concepts
* Moles: We'll use moles to represent the amounts of reactants and products. One mole of a substance contains Avogadro's number of particles (6.022 x 10^23).
* Molar Mass: The molar mass of a substance is the mass of one mole of that substance.
* Stoichiometry: This is the branch of chemistry that deals with the relative amounts of reactants and products in chemical reactions.
Calculations
1. Calculate Moles:
* Fluorine (F₂):
* Molar mass of F₂ = 38 g/mol (19 g/mol per fluorine atom x 2 atoms)
* Moles of F₂ = 19 g / 38 g/mol = 0.5 moles
* Bromine (Br₂):
* Molar mass of Br₂ = 160 g/mol (80 g/mol per bromine atom x 2 atoms)
* Moles of Br₂ = 180 g / 160 g/mol = 1.125 moles
2. Determine Moles of Reactants and Products:
* Hydrogen Fluoride (HF):
* From the balanced equation, 1 mole of F₂ reacts to produce 2 moles of HF.
* Moles of HF = 0.5 moles F₂ x (2 moles HF / 1 mole F₂) = 1 mole HF
* Hydrogen Bromide (HBr):
* From the balanced equation, 1 mole of Br₂ reacts to produce 2 moles of HBr.
* Moles of HBr = 1.125 moles Br₂ x (2 moles HBr / 1 mole Br₂) = 2.25 moles HBr
3. Calculate Mass of Products:
* Hydrogen Fluoride (HF):
* Molar mass of HF = 20 g/mol (1 g/mol for hydrogen + 19 g/mol for fluorine)
* Mass of HF = 1 mole HF x 20 g/mol = 20 g
* Hydrogen Bromide (HBr):
* Molar mass of HBr = 81 g/mol (1 g/mol for hydrogen + 80 g/mol for bromine)
* Mass of HBr = 2.25 moles HBr x 81 g/mol = 182.25 g
Summary
* Hydrogen Fluoride (HF):
* 0.5 moles of F₂ react with 0.5 moles of H₂ to produce 1 mole (20 g) of HF.
* Hydrogen Bromide (HBr):
* 1.125 moles of Br₂ react with 1.125 moles of H₂ to produce 2.25 moles (182.25 g) of HBr.
Important Note: These calculations assume that hydrogen is in excess, meaning there is enough hydrogen to react completely with the given amounts of fluorine and bromine.
