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Answer :

Dsdrajlingo

Answer:

62.9%

Explanation:

Step 1: Given data

  • Mass of hydrate sodium carbonate (mNa₂CO₃.xH₂O): 71.5 g
  • Mass of anhydrous sodium carbonate (mNa₂CO₃): 26.5 g

Step 2: Calculate the mass lost of water

We will use the following expression.

mH₂O = mNa₂CO₃.xH₂O - mNa₂CO₃

mH₂O = 71.5 g - 26.5 g = 45.0 g

Step 3: Calculate the percent of water in the hydrate sodium carbonate

We will use the following expression.

%H₂O = mH₂O / mNa₂CO₃.xH₂O × 100%

%H₂O = 45.0 g / 71.5 g × 100%

%H₂O = 62.9%

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