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ale4655 [162]
3 years ago
10

With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?

Chemistry
1 answer:
mina [271]3 years ago
8 0

Answer: 7.07 grams

Explanation:

To calculate the moles :

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

\text{Moles of} zinc=\frac{21g}{65g/mol}=0.32moles

\text{Moles of} CuCl_2=\frac{7g}{134g/mol}=0.052moles

Zn+CuCl_2\rightarrow Cu+ZnCl_2

According to stoichiometry :

1 mole of CuCl_2 require 1 mole of Zn

Thus 0.052 moles of CuCl_2 will require=\frac{1}{1}\times 0.052=0.052moles  of Zn

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

As 1 mole of CuCl_2 give = 1 mole of ZnCl_2

Thus 0.052 moles of CuCl_2 give =\frac{1}{1}\times 0.052=0.052moles  of ZnCl_2

Mass of ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g

Thus 7.07 g of ZnCl_2 will be produced from the given masses of both reactants.

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The final temperature is about 39.1 °C.

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