Need Help On This Question.!
Potassium chloride reacts with bromine forming potassium bromide plus chlorine. If 500g of potassium bromide are produced, how many grams of potassium chloride reacted?
Flash! I have news for you. Chlorine will replace Br^- from a salt; however, Br2 will NOT displace Cl^- from a salt.
This will occur.
Cl2 + 2KBr ==> Br2 + 2KCl but the reverse will NOT occur.
To determine the grams of potassium chloride that reacted, we need to use stoichiometry, which is based on the balanced chemical equation for the reaction. Let's first write down the balanced equation for the reaction:
2KCl + Br2 → 2KBr + Cl2
From the balanced equation, we see that 2 moles of potassium chloride react with 1 mole of bromine to produce 2 moles of potassium bromide and 1 mole of chlorine.
Step 1: Calculate the molar mass of potassium bromide (KBr).
The molar mass of KBr is 119 g/mol (39 g/mol for potassium + 80 g/mol for bromine).
Step 2: Convert the given 500 grams of potassium bromide into moles.
moles = mass / molar mass
moles of KBr = 500 g / 119 g/mol ≈ 4.20 moles
Step 3: Use stoichiometry to determine the number of moles of KCl that reacted.
From the balanced equation, we can see that 2 moles of KBr are produced for every 2 moles of KCl that react.
So, the number of moles of KCl that reacted = 2.10 moles.
Step 4: Convert the moles of KCl to grams.
mass = moles × molar mass
mass of KCl = 2.10 moles × 74.6 g/mol (molar mass of KCl)
mass of KCl = 156.66 g
Therefore, approximately 156.66 grams of potassium chloride reacted.