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3 years ago

How many moles of chlorine are used up in a reaction that produces 0.35kg of BCl3

Chemistry

1 answer:

maw [93]3 years ago

4 0

4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.

(a) The <em>balanced chemical equation </em>is

2B + 3Cl2 → 2BCl3

(b) Convert kilograms of BCl3 to moles of BCl3

MM: B = 10.81; Cl = 35.45; BCl3 = 117.16

Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3

Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2

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2 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

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Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

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Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

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Therefore, the percent abundance of the heaviest isotope is, 78 %

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3 years ago

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2 years ago

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