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Answer:
All the amounts of reactants and products are:
KBr
11.0g (given)
0.0924 mol
Cl₂
0.0462 mol
3.28g
KCl
0.0924 mol
6.89 g
Br₂
0.0462 mol
7.39 g
Explanation:
<u>1. Balanced chemical equation (given)</u>
<u>2. Mole ratios</u>
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<u>3. Molar masses</u>
- Molar mass Cl₂: 70.906g/mol
- Molar mass KBr: 119.002 g/mol
- Molar mass KCl: 74.5513 g/mol
- Molar mass KBr: 159.808 g/mol
<u>4. Convert 11 grams of potassium bromide to moles:</u>
- #moles = mass in grams / molar mass
- #mol KBr = 11g / 119.002g/mol = 0.092435mol KBr
<u>5. Use the mole ratios to find the amounts of Cl₂, KCl, and Br₂</u>
a) Cl₂
b) KCl
c) Br₂
The final calculations are rounded to 3 sginificant figures.