How do you balance $(NH_4)_2CO_3 -> NH_3 + CO_2 + H_2O$ ?

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Answer 1 · 2016-03-15T22:22:58

This equation can be balanced using trial and error method. The most common method of balancing chemical equations.

$(NH_4)_2CO_3 -> 2NH_3 + CO_2 + H_2O$

First is to account the initial number of atoms as reference, both the reactant and the products using the T balance;
Present the elements according to its appearance in the reactant side and copy the same in the product side; though it's no rule on this;
$Reactant TTProduct$
$N=2$ ...........l $N=cancel1 2$
$H=8$ ...........l $H=cancel5 8$
$C=1$ ............l $C=1$
$O=3$ ............l $O=3$
Generally, complex compounds are always considered first and work your way to the least one;
Every change you make with coefficients, make sure to recount the number of atoms of the involve elements and record using your T balance and of course cancelling the previous counts;
In this case, putting 2 as coefficient in the first product of the reaction will suffice the equation.
The balanced equation is
$(NH_4)_2CO_3->2NH_3+CO_2+H_2O$

How do you balance (NH_4)_2CO_3 -> NH_3 + CO_2 + H_2O(NH_4)_2CO_3 -> NH_3 + CO_2 + H_2O?