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lesya692 [45]
3 years ago
14

Given the following masses, calculate the mass of water lost & determine the percent hydration.

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
6 0
<span>You are given the mass of hydrate which is 10.24g and the mass of anhydrate which is 8.74g. Hydrate means that it contains water and anhydrate means the substance has lost water during the process. So you just subtract them, 10.24g – 8.74g = 1.5g of water is lost. Percent hydration is equal to one subtracted to the mass of anhydrate divided by the mass of hydrate multiplied by 100. So we have, [1-(8.74g/10.24g)] *100 = 14.65% or 15%. Your answers are correct.</span>
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Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

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To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

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