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2 years ago

Zn-64 = 48.63%

Chemistry

1 answer:

sladkih [1.3K]2 years ago

6 0

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

Zn-64 = 48.63%

Zn-66 = 27.90%

Zn-67 = 4.10%

Zn-68 = 18.75%

Zn-70 = .62%

We need to find the average atomic mass of the given element. It can be solved as follows :

$A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u$

So, the average atomic mass of zinc is 65.46 u.

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The maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

<h3>Stoichiometry </h3>

From the question, we are to determine the maximum mass of anhydrous zinc chloride that could be obtained

From the given balanced chemical equation,

ZnO + 2HCl → ZnCl₂ + H₂O

This means 1 mole of ZnO will completely react with 2 moles of HCl to produce 1 mole of ZnCl₂ and 1 mole of H₂O

From the given information

Number of moles of ZnO = 0.0830 mole

Now, we will calculate the number of moles of HCl that is present

Volume of HCl added = 100 cm³ = 0.1 dm³

Concentration of the HCl = 1.20 mol/dm³

Using the formula,

Number of moles = Concentration × Volume

Number of moles of HCl present = 1.20 × 0.1 = 0.120 mole

Since

1 mole of ZnO will completely react with 2 moles of HCl to produce 1 mole of ZnCl₂

Then,

0.06 mole of ZnO will react with the 0.120 mole of HCl to produce 0.06 mole of ZnCl₂

Therefore, maximum number of moles of anhydrous zinc chloride that could be produced is 0.06 mole

Now, for the maximum mass that could be produced

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of ZnCl₂ = 136.286 g/mol

Then,

Mass = 0.06 × 136.286

Mass = 8.17716 g

Mass ≅ 8.18 g

Hence, the maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

Learn more on Stoichiometry here: brainly.com/question/11910892

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There are four different starting molecules that one might use to synthesize the illustrated alkyl halide as the major product u

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Answer:

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