Anonymous
Anonymous asked in Science & MathematicsChemistry · 6 years ago

How to complete a chemical equation?

Hi I need to know how to complete a chemical equation such as

FeCl2 + Na3PO4 =

Sr + Pb(NO3)2 =

NaOH + H3PO4 =

Fe(C2H3O2)3 + HCl =

And so on...

I missed the lesson in class and could use some help on this paper. Thanks :)

Update:

Ba N2 =

Bi2O3 H2 =

Ag2O =

CrF3 Br2

1 Answer

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  • Anonymous
    6 years ago
    Best Answer

    There are two steps in balancing a chemical equation:

    1. Balancing the two sides elemental (or functional group) composition.

    2. Balancing the gain and loss of electrons. (Balancing charge).

    Step 2 is not the same (generally) as balancing the stoichiometry so that charges are balanced for the various species present.

    For instance Fe(+3) + Cl(-) is charged balanced in FeCl₃

    while 2Fe(+3) + H₂ → 2Fe(+2) + 2H(+) is also charged balanced but in addition the ferric iron has gained one electron (each) and has been reduced, while the molecular hydrogen has been oxidized from H° to H(+). Normally, the gain and loss of electrons (redox (aka oxidation/reduction) equations) are a FAR more advanced topic, and I would assume you don't have to worry about it yet, but the second of your equations has Sr which I assume is Sr° (or you should be ashamed of yourself for not even understanding how to write chemical species!). So that reaction is a redox reaction.

    Nicely, each Sr will loose 2 electrons which will allow each Pb atom to gain two, so that the reaction is Sr° + Pb(NO₃)₂ → Sr(NO₃)₂ + Pb° the loss/gain of electrons balance, and it should be obvious that the number of atoms of each element are the same on LHS and RHS.

    -=-=-

    All of these reactions are ionic, meaning that you need to understand that the groups of cations are the metals and H and the group of anions are the Chloride ion and the other polyatomic ions (PO₄(-3) , etc.). You just need to learn the common ions and their charges. In this particular problem, all you need to know is that H usually is the H(+) ion (sometimes rarely it is H° or H(-), but that is another two advanced subjects).

    You need to know that Na is either Na° (the metal) or Na(+) the ion, and that Cl is almost always Cl(-) { or Cl° in Cl₂, chlorine gas}. Using these three rules of thumb, you can balance charges and atoms on the RHS of the equations.

    ---------------------

    So to start out you need only worry about Step 1 above.

    First identify the ions present. Then identify their charge.

    Assume that their charge will remain the same on the RHS (right hand side). Count up the number of ions (or atoms, as appropriate) on the LHS. On the right, write the formulas of each product compound, making sure that they are charge neutral (cations balance anions).

    Once that is done, you need to figure out the coefficients before each species so that there are the same number of atoms of each type on the RHS as there are on the LHS. There is a mathematical way to do this, but it is far too complex for every day use, and you just need to learn how to do it in your head.

    examples:

    A + B → A₂B tells you that the LHS needs at least 2 A's

    different example:

    A + B₂ → A₂B tells you that the RHS needs 2 A₂B's which in turn tells you that you need 4 A's ...final equation 4A+B₂ →2A₂B

    -=-=-=-

    I am assuming that you know how to count up atoms?

    2ABC means two of each A, B, & C

    A(BC)₂ means one A and two of each B and C.

    2A(BC)₂ means two A's and 2x2 B's and C's = 4 each.

    And A₂(B₃(CD₂)₂)₄ is 2A, 12B and 8C and 16D ...etc. etc.

    The only other thing you will see from time to time is something like this"

    AB•5C which means the C is a separate compound (usually) but in this case is part of the structure of the species AB•5C and there are 5 of them. CuSO₄•5H₂O is the common example. The 5 H₂O molecules (water) are "water of crystallization" and part of the crystal structure.

    You treat the leading 5 as you would in front of any compound.

    I won't explain how to do Step 2. As said, an advanced subject (but not rocket science, either, lol)

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