a hydrate of pottasium carbonate has the formula K2Co3.xH2O. A 10.00g sample of the hydrated solid is heated, and forms 7.93g of anhydrous salt. Calculate moles of water, moles of anhydrous salt, formula of hydrate.
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Explanation:
It is not too difficult. First, calculate the number of moles of anhydrous salt are present in 7.93 g. To do this you need to determine the molecular weight of K2CO3. This is 2x39.1 + 12 + 3x16 = 138.2 g/mole. So:
The number of moles of K2CO3 is equal to 7.93 g/138.2 g/mole = 0.0574 mole.
The loss of water amounts to 10 g - 7.93 g = 2.07 grams of water. Given that the molecular weight of water (H2O) is 2x1 + 16 = 18 g/mole, the number of moles that were lost is:
2.07 g / 18 g/mole = 0.115 mole.
The ratio of water to K2CO3 (which is what ‘x’ means in K2CO3xH2O) is equal to:
0.115 mole of water/0.0574 mole of potassium carbonate = 2
So, x is 2 and the formula is K2CO3.2H2O.
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