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Kipish [7]
3 years ago
11

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?
Chemistry
1 answer:
sergejj [24]3 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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