How many moles are present in 63.80L of oxygen gas, O2?
1 answer:
Answer:
2.85moles of oxygen gas
Explanation:
Given parameters:
Volume of oxygen gas = 63.8L
Unknown:
Number of moles = ?
Solution:
We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:
1 mole of a gas at STP occupies a volume of 22.4L
So;
63.8L of oxygen gas will take up a volume of
= 2.85moles of oxygen gas
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14 grams of calcium oxide.
Thermal decomposition of 25 grams of calcium carbonate would result in the production of 14 grams of calcium oxide.
We know that,
CaCO₃ → CaO + CO₂
First, the following quantities react and are generated according to the stoichiometry of the reaction, which is the relationship between the amounts of reagents and products in a chemical reaction:
- CaCO₃: 1 mole
- CaO: 1 mole
- CO₂: 1 mole
Being:
- Ca: 40 g/mole
- C: 12 g/mole
- O: 16 g/mole
- CaCO₃: 40 g/mole + 12 g/mole + 3*16 g/mole= 100 g/mole
- CaO: 40 g/mole + 16 g/mole= 56 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
The following mass amounts of the compounds involved in the reaction then react and are created, according to the reaction's stoichiometry:
- CaCO₃: 1 mole* 100 g/mole= 100 g
- CaO: 1 mole* 56 g/mole= 56 g
- CO₂: 1 mole* 44 g/mole= 44 g
How much calcium oxide will be produced if, according to the stoichiometry of the reaction, 100 grams of calcium carbonate CaCO3 result in 56 grams of calcium oxide CaO and 25 grams of CaCO3?
mass of calcium oxide =
mass of calcium oxide= 14 grams
14 grams of calcium oxide would be produced by thermal decomposition of 25 grams of calcium carbonate.
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