Ask Question
21 March, 09:45

A hydrate of beryllium nitrate has the following formula: Be (NO3) 2⋅xH2O. The water in a 3.41-g sample of the hydrate was driven off by heating. The remaining sample had a mass of 2.43 g. Find the number of waters of hydration (x) in the hydrate.

+4
Answers (1)
  1. 21 March, 09:57
    0
    The formula is Be (NO3) 2*3H2O. The number of waters in the hydrate is 3

    Explanation:

    Step 1: Data given

    Mass of the hydrate sample = 3.41 grams

    Mass after heating = 2.43 grams

    Step 2: Calculate mass of water

    After heating, all the water is gone. So the mass of water can be calculated by

    Mass of hydrate before heating - mass after heating

    Mass of water = 3.41 - 2.43 = 0.98 grams

    Step 2 : Calculate moles H2O

    Moles H2O = mass H2O / molar mass H2O

    Moles H2O = 0.98 grams / 18.02 g/mol

    Moles H2O = 0.054 moles

    Step 3: Calculate moles Be (NO3) 2

    Moles Be (NO3) 2 = 2.43 grams / 133.02

    Moles Be (NO3) 2 = 0.0183 moles

    Step 4: Calculate molecules water

    Molecules H2O = 0.054 moles / 0.0183 moles

    Molecules H2O = 3.0

    The formula is Be (NO3) 2*3H2O. The number of waters in the hydrate is 3
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A hydrate of beryllium nitrate has the following formula: Be (NO3) 2⋅xH2O. The water in a 3.41-g sample of the hydrate was driven off by ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers