How many grams of water are needed to form 21.8 grams of oxygen gas?
1 answer:
Answer:
24.5g
Explanation:
Given parameters:
Mass of oxygen gas = 21.8g
Unknown:
Mass of water = ?
Solution:
This problem deals with the decomposition of water;
Reaction equation;
2H₂O → 2H₂ + O₂
2 mole of water will produce 1 mole of oxygen gas
First find the number of mole of the oxygen gas given;
Number of moles =
molar mass of O₂ = 2(16) = 32g/mol
Number of moles of O₂ = = 0.68moles
1 mole of oxygen gas will be produced from 2 mole of water
0.68moles of oxygen gas will be produce from 0.68 x 2 moles of water; = 1.36moles of water
Mass of water = number of moles x molar mass
Molar mass of water = 2(1) + 16 = 18g/mol
Mass of water = 1.36 x 18 = 24.5g
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2) So, one alkyl group is CH3 and second one can be CH or CH2.
3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.
4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.
The ratio remains the same, 3:2 ie 6:4
