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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)​

Respuesta :

Neetoo

Answer:

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

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.

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