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aleksandr82 [10.1K]
4 years ago
6

On heating 5.03 g of hydrated barium chloride (BaCl2 XH2O), 4.23 g is left behind. What is the name of this hydrate? (

Chemistry
1 answer:
RSB [31]4 years ago
8 0

<span>1)    </span><span>Deduce the two masses and see the amount of water was driven off when heated: </span><span>

<span>5.03 g - 4.23 g = 0.8 g H2O given off </span>

<span>2) Change mass from grams to moles of H2O: </span>

<span>0.8 g H2O / 18 g H2O in 1 mole = 0.044 mol H2O </span>

<span>3) Change left over mass to moles  of BaCl2 .</span></span>

<span>
<span>4.23 g BaCl2 / 207 g BaCl2 in 1 mol = 0.021 mol BaCl2 </span>

<span>4)Find the ratio of mol H2O to mol BaCl2: </span>

<span>0.044 mol H2O : 0.021 mol BaCl2 </span>

<span>5) The resulting  ratio is  2:1 so two H2O for each BaCl2, thus,  the hydrate was named: </span>
<span>Barium chloride di-hydrate</span></span>

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Valence shell brainly.com/question/10901976

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