Calculate the percentage of water of hydration in barium chloride crystals (the atomic weight of Ba is about 137.4 g/mol)?

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Calculate the percentage of water of hydration in barium chloride crystals (the atomic weight of Ba is about 137.4 g/mol)?

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Answer 1 · 2018-07-15T04:45:43

14.75%

Explanation:

You'll need to start off by knowing the formula of the barium chloride crystals, which is barium chloride dihydrate, $B a C l_{2} \cdot 2 H_{2} O$ $BaCl_2 *2H_2O$ . From the formula, you'll notice that there are 2 moles of $H_{2} O$ $H_2O$ for every mole of the hydrate, $B a C l_{2} \cdot 2 H_{2} O$ $BaCl_2 *2H_2O$ .

When the 1 mole of hydrate crystals are heated, you will be left with 1 mole of the anhydrous crystal ( $B a C l_{2}$ $BaCl_2$ ) and 2 moles of $H_{2} O$ $H_2O$ , like this:

$B a C l_{2} \cdot 2 H_{2} O \to B a C l_{2} + 2 H_{2} O$ $BaCl_2 *2H_2O -> BaCl_2 + 2 H_2O$

To calculate the percentage of water of hydration, we'll need to find the molar mass of water ( $H_{2} O$ $H_2O$ ) and hydrate crystal ( $B a C l_{2} \cdot 2 H_{2} O$ $BaCl_2 *2H_2O$ ).

$molar mass H_{2} O = 2 (1.01) + 16.00 = 18.02 g/mol$ $"molar mass " H_2O = 2(1.01) + 16.00 = 18.02 " g/mol"$

$molar mass B a C l_{2} \cdot 2 H_{2} O = 137.4 + 2 (35.45) + 2 [2 (1.01) + 16.00] = 244.34 g/mol$ $"molar mass " BaCl_2 *2H_2O = 137.4 + 2(35.45) + 2[2(1.01) + 16.00] = 244.34 " g/mol"$

We can then proceed to calculating the percentage of water of hydration in barium chloride crystals:
$Percentage of water of hydration = \frac{2 \times 18.02}{244.34} \cdot 100 % = 14.75 %$ $"Percentage of water of hydration" =(2 xx 18.02 ) / (244.34)*100%=14.75%$

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Calculate the percentage of water of hydration in barium chloride crystals (the atomic weight of Ba is about 137.4 g/mol)?

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