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A 118.63 g sample of SrCl2·6H2O was heated and 34.81 grams of water were recovered from the sample. What percent of the original H2O remains in the sample following this heating?

User Pavunkumar
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1 Answer

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Percentage mass of water in SrCl₂·6H₂O
= (6 x 18) / (88 + 35.5 x 2 + 6 x 18)
= 40.4%

Mass of water in sample originally₂:
118.63 x 40.4%
= 47.9 grams

Percentage remaining = (47.9 - 34.81) / 47.9 x 100
= 27.3%
User Ivstas
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