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2 years ago

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how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas if t

he reaction goes to completion?

1 answer:

sergejj [24]2 years ago

8 0

Answer:

27984.1311 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

<u>For calcium :- </u>

Mass of calcium = 20.0 kg = 20000 g ( 1 kg = 1000 g )

Molar mass of calcium = 40.078 g/mol

Thus,

$Moles= \frac{20000\ g}{40.078\ g/mol}$

$Moles\ of\ calcium= 499.0269\ mol$

<u>For oxygen gas :- </u>

Mass of oxygen gas = 20.0 kg = 20000 g ( 1 kg = 1000 g )

Molar mass of oxygen gas = 31.999 g/mol

Thus,

$Moles= \frac{20000\ g}{31.999\ g/mol}$

$Moles\ of\ oxygen\ gas= 625.0195\ mol$

According to the given reaction:

$2Ca+O_2\rightarrow 2CaO$

2 moles of calcium reacts with 1 moles of oxygen gas

Also,

1 mole of calcium react with 1/2 mole of oxygen gas

So,

499.0269 mole of calcium react with $\frac{1}{2}\times 499.0269$ mole of oxygen gas

Moles of oxygen gas = 249.5135 moles

Available moles of oxygen gas = 625.0195 moles

Limiting reagent is the one which is present in small amount. Thus, calcium is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

2 moles of calcium produces 2 moles of calcium oxide

So,

1 mole of calcium produces 1 mole of calcium oxide

Also,

499.0269 mole of calcium produces 499.0269 mole of calcium oxide

Moles of calcium oxide = 499.0269 moles

Molar mass of calcium oxide = 56.0774 g/mol

Mass of calcium oxide = Moles × Molar mass = 499.0269 × 56.0774 g = 27984.1311 g

<u>27984.1311 g of calcium oxide will be produced.</u>

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Explanation:

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Using the statement of this problem, we can deduce that;

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Answer:

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Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

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4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

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2 years ago

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2 years ago