Question

How many grams of oxygen are produced if you have ${{10}}{{.0}}$ grams of water available for the following reaction?
${{2}}{{{H}}_{{2}}}{{O}} \to {{2}}{{{H}}_{{2}}}{{ + }}{{{O}}_{{2}}}$

Answer 1

In the above question, since we are provided with balance equation of dissociation of water molecule and we can infer from it that 2 moles of water produces one mole of oxygen molecule and hence, first we have find the number of moles of water present and then we can find the number of moles of oxygen molecule present and hence, the weight of oxygen can be obtained.

Formula Used
${{n = }}\dfrac{{{m}}}{{{M}}}$
Where n is the number of moles
m = given mass
M = molar mass.

Complete step by step solution
In the above question, the weight of water present is given as 10g.
So, let us now find out the number of moles of water present. For this we have to find out the molar mass of water.
Molar mass of ${{{H}}_{{2}}}{{O}}$ = 2 ${{ \times }}$ atomic mass of H + atomic mass of O = ${{2 \times 1 + 16 = 18}}$ g
${{n = }}\dfrac{{{m}}}{{{M}}}{{ = }}\dfrac{{{{10}}}}{{{{18}}}}{{ = 0}}{{.55}}$
Hence, ${{0}}{{.55}}$ moles of water is present.
We have a balanced chemical equation as:
${{2}}{{{H}}_{{2}}}{{O}} \to {{2}}{{{H}}_{{2}}}{{ + }}{{{O}}_{{2}}}$
Here, we can see that 2 moles of water form 1 mole of oxygen.
${{0}}{{.55}}$ moles of water form $\dfrac{{{1}}}{{{2}}}{{ \times 0}}{{.55 = 0}}{{.27}}$ moles of oxygen.
Let us now find out molar mass of oxygen:
Molar mass of ${{{O}}_{{2}}}$ = 2 ${{ \times }}$ atomic mass of O = ${{2 \times 16 = 32}}$ g
We know that ${{n = }}\dfrac{{{m}}}{{{M}}}$
So, by rearranging, we get:
${{m = n \times M = 0}}{{.27 \times 32 = 8}}{{.88}}$ g
Therefore, ${{8}}{{.88}}$ g of oxygen is formed when 10g of water is present.

Note
1 mole = $6.022 \times {10^{23}}$ atoms, or molecules, or protons, or electrons etc.
In short, it is the number of particles in a mole.
A reaction equation which is not balanced is also known as a skeletal equation.