Question about Moles of a Compound in a Balanced Chemical Equation  [ENDORSED]

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Prabhdeep-3L
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Joined: Sat Jul 22, 2017 3:00 am

Question about Moles of a Compound in a Balanced Chemical Equation

Postby Prabhdeep-3L » Sat Oct 07, 2017 4:49 pm

There was an example in the lecture about Balancing Chemical Equations that involved the balanced equation of:

2Na + 2H2O -> 2NaOH + H2

and then showed

2 moles + 2 moles -> 2 moles + 1 mole
underneath it.

4 moles isn't equal to 3 moles (at least not in the way I'm thinking about it) so I don't understand how it's related to the balanced chemical equation where according to the law of the conservation of mass, the reactants are equal to the products.
So my ultimate question is: What correlation do the moles of the compounds have with the balanced chemical equation?

Eryn Wilkinson 3H
Posts: 20
Joined: Fri Sep 29, 2017 7:07 am

Re: Question about Moles of a Compound in a Balanced Chemical Equation  [ENDORSED]

Postby Eryn Wilkinson 3H » Sat Oct 07, 2017 5:01 pm

The mass on both sides of the equation has to be equal, but not the number of moles (like on the test when it asked for the net number of moles produced). The coefficients balance the equation because the number of atoms of each element on each side has to be the same, so if there are 2 moles of Na on the reactants side there must be 2 on the products side. However, when the reaction occurs new compounds form, and even though there may be fewer moles of a particular compound it has a greater molar mass.

StephanieRusnak
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Re: Question about Moles of a Compound in a Balanced Chemical Equation

Postby StephanieRusnak » Sat Oct 07, 2017 7:38 pm

Mass is still conserved because there is the same number of atoms on both sides of the equation. Moles can be produced/lost but mass cannot

Abdullah2I
Posts: 41
Joined: Fri Sep 29, 2017 7:06 am

Re: Question about Moles of a Compound in a Balanced Chemical Equation

Postby Abdullah2I » Sat Oct 07, 2017 7:47 pm

Yeah, I was really confused about this as well, but this helps. Thanks!


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