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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.

Answer :

Answer:

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

The number of waters of hydration in the hydrate is 3.

Based on the given information,  

• The formula of the beryllium nitrate hydrate is Be(NO3)2⋅xH2O.  

• The weight of the sample is 3.41 grams.

• The weight of the sample after heating of the sample is 2.43 grams {weight of Be(NO3)2}.  

Now, the mass of xH2O will be,  

= 3.41 g - 2.43 g  

= 0.98 grams

The moles of Be(NO3)2 will be calculated as,  

Mass of Be(NO3)2/Molecular weight of Be(NO3)2 = 2.43 g/133 = 0.01827 moles

The moles of xH2O = 0.98/18 = 0.05444 moles

Now the value of x will be,  

= 0.05444/0.01827  

= 3

Thus, the number of waters of hydration in the hydrate is 3.

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