Balancing a chemical equation with multiple compounds

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Ryan Cerny 3I
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Balancing a chemical equation with multiple compounds

Postby Ryan Cerny 3I » Mon Sep 26, 2016 11:16 pm

Here is a long and complicated trick I can up with balance a complex chemical equation:
1) Find an atom this is only contained in one reactant and in one product. For which ever molecule contains more of that atom, assign a coefficient of "X" and scale the the molecules' coefficient (with respect to "X") according to how much of that atom is in that other compound.
Ex: Balancing: _C10H15N + _O2 --> _CO2 + _H2O + _CH4N2O
_C10H15N + _O2 --> _CO2 + _H2O + CH4N2O; chose N
2XC10H15N + _O2 --> _CO2 + _H2O + XCH4N2O; the last product has more N than does the first reactant, therefore we assign it a coefficient of X and scale the other one accordingly.
2) Find a different atom that is not only composed in these two compounds (assigned respective "X" coefficients) but is also in a 3rd compound. Put a "Y" infront of that 3rd compound. Multiply the variable coefficients by the number of the new atoms for all three. Add the products on the respective sides and set the multiplied units of the reactants equal to that of the products; solve for "Y" with respect to "X."
2XC10H15N + _O2 --> _CO2 + _H2O + XCH4N2O; chose C it is not only composed in C10H15N and CH4N2O, but also in C02.
2Xx10 (C atoms) = Yx1 (C atom) + Xx4 (C atoms); Y = 19X
2XC10H15N + _O2 --> 19XCO2 + _H2O + XCH4N2O
3) And Finish the problem with the same logical approach.
2XC10H15N + 26XO2 --> 19XCO2 + 13XH2O + XCH4N2O; X=1.
Balanced equation: 2C10H15N + 26O2 --> 19CO2 + 13H2O + CH4N2O.

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Re: Balancing a chemical equation with multiple compounds

Postby Chem_Mod » Tue Sep 27, 2016 9:22 am


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