Define the term standardization
1 answer:
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Answer:
27984.1311 g
Explanation:
The formula for the calculation of moles is shown below:
<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,
<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,
According to the given reaction:
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 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>