Question

How many grams of K2CO3 (MM = 138.21 g/mol)are needed to completely react with 453.2 mL of

0.525 M HBr?
2 HBr(aq) + K2CO3(s) → 2 KBr(aq) + CO2(g) + H20(1)​

Answer 1

16.45 g of K₂CO₃ are required to react completely with 0.525 M HBr.

Step-by-step explanation:

Given data:

Volume of HBr solution = 453.2 mL (453.2 /1000 = 0.4532 L)

Molarity of HBr = 0.525 M

Molar mass of K₂CO₃ = 138.21 g/mol

Mass of K₂CO₃ = ?

Solution:

Chemical equation:

2HBr + K₂CO₃ → 2KBr + CO₂ + H₂O

Number of moles of HBr:

Molarity = moles of solute / volume in litter

0.525 M = moles of solute / 0.4532 L

Moles of solute = 0.525 g/L ×0.4532 L

Moles of solute = 0.238 mole

Now we compare the moles of HBr and K₂CO₃ from balance chemical equation:

HBr : K₂CO₃

2 : 1

0.238 : 1/2× 0.238 = 0.119 moles

Mass of K₂CO₃:

Mass of K₂CO₃ = moles × molar mass

Mass of K₂CO₃ = 0.119 mol × 138.21 g/mol

Mass of K₂CO₃ = 16.45 g

16.45 g of K₂CO₃ are required to react completely with 0.525 M HBr.