asked 222k views
5 votes
If 5.4 moles of Na₂CO₃ react with excess calcium hydroxide. how many grams of CaCO₃ will be produced?

Na₂CO₃+Ca(OH)₂=2NaOH+CaCO₃

asked
User Pengson
by
7.0k points

1 Answer

2 votes

Answer:

540.47g approximately

Step-by-step explanation:

No. of moles in Na₂CO₃ = 5.4 moles

Mole ratio of Na₂CO₃ : CaCO₃ = 1:1

No. of moles in CaCO₃ = 5.4 moles

Mass of CaCO₃ = 5.4 × 100.0869

= 540.46926g

answered
User Maligree
by
8.9k points
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