Question

ok, now that you know what a balanced chemical equation looks like and how to balance a more complicated equation, lets take this one step further. lets try to balance an equation with polyatomic ions. remember, for polyatomic ions, the number outside of the parenthesis acts like a multiplier. so, in a compound of $ca(no_{3})_{2}$ there would be... 1 calcium atom, 2 nitrogen atoms, and 6 oxygen atoms. if you were to add a stoichiometric coefficient of two before such a compound such as $2ca(no_{3})_{2}$, there would be 2 calcium, 4 nitrogen, 12 oxygen atoms. (see image on the right) use this information to balance the equation below. $sc_{2}o_{3}+h_{2}o
ightarrow sc(oh)_{3}$

Explanation:

Step1: Balance Sc atoms

On the left - hand side, in \(Sc_2O_3\), there are 2 Sc atoms. On the right - hand side, in \(Sc(OH)_3\), let the coefficient of \(Sc(OH)_3\) be \(x\). For Sc atoms: \(2 = x\). So we put a coefficient of 2 in front of \(Sc(OH)_3\), and the equation becomes \(Sc_2O_3+H_2O
ightarrow2Sc(OH)_3\).

Step2: Balance O atoms

In \(Sc_2O_3\), there are 3 O atoms. In \(2Sc(OH)_3\), the number of O atoms from \(Sc(OH)_3\) is \(2\times3 = 6\) (from the \(OH\) groups). Let the coefficient of \(H_2O\) be \(y\). The total number of O atoms on the left - hand side is \(3 + y\) (3 from \(Sc_2O_3\) and \(y\) from \(H_2O\)), and on the right - hand side is 6 (from \(2Sc(OH)_3\)). So \(3 + y=6\), which gives \(y = 3\). The equation now is \(Sc_2O_3 + 3H_2O
ightarrow2Sc(OH)_3\).

Step3: Balance H atoms

On the left - hand side, in \(3H_2O\), there are \(3\times2=6\) H atoms. On the right - hand side, in \(2Sc(OH)_3\), there are \(2\times3 = 6\) H atoms.

Answer:

\(1Sc_2O_3+3H_2O
ightarrow2Sc(OH)_3\)