How many electrons are transferred in the production of 6.06×10^−4 mol H2(g) from water?
Two electrons per water molecule, or two electrons per H2 molecule.
numberelectrons= 2moleselctrons/moleH2*6.06E-4 molesH2
thank you :)
To determine the number of electrons transferred in the production of H2(g) from water, we need to understand the chemical reaction involved.
The balanced equation for the production of H2(g) from water is:
2H2O(l) -> 2H2(g) + O2(g)
From the equation, we can see that for every 2 molecules of water, 2 molecules of H2(g) (hydrogen gas) and 1 molecule of O2(g) (oxygen gas) are produced.
Now, we need to consider that 6.06x10^(-4) mol of H2(g) is produced from water. From the above equation, we know that 2 moles of H2(g) are produced for every 2 moles of water.
Therefore, we can say that for every 2 moles of H2(g) produced, 2 moles of electrons are transferred.
To find the number of electrons transferred for 6.06x10^(-4) mol of H2(g), we can set up a proportion:
2 moles H2(g) / 2 moles electrons = 6.06x10^(-4) moles H2(g) / n moles electrons,
where n represents the number of electrons transferred.
By cross-multiplying and solving for n, we find:
2 moles H2(g) * n moles electrons = 2 moles electrons * 6.06x10^(-4) moles H2(g)
n = (2 moles electrons * 6.06x10^(-4) moles H2(g)) / 2 moles H2(g)
Simplifying the equation gives:
n = (2 * 6.06x10^(-4)) / 2
n = 1.212x10^(-4)
Therefore, approximately 1.212x10^(-4) moles of electrons are transferred in the production of 6.06x10^(-4) mol of H2(g) from water.