How many grams of water are needed to form 21.8 grams of oxygen gas?

Chemistry

1 answer:

GaryK [48]2 years ago

5 0

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 = $\frac{mass}{molar mass}$

molar mass of O₂ = 2(16) = 32g/mol

Number of moles of O₂ = $\frac{21.8}{32}$ = 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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