Question

How do you write the balanced equation for strontium hydroxide and zirconium (I) perchlorate

Answer 1

The law of conservation of mass states that the mass of a system must remain constant everytime, it can neither be created or destroyed. This means the number of atoms of each element on reactant side must equal the number of atoms of each element on product side.

First we figure out the chemical formula for each compound by taking advantage of oxidation states which are able to tell us whether an element will lose or gain an electron

Strontium hydroxide?

Sr is an alkaline earth metal which has 2 valence electrons that when both are lost it has an oxidation state of +2. Since we know the hydroxide ion has a charge of -1, $-_{OH}$, Then two $-_{OH}$ are needed to form compound, hence we have $SrOH_2$

zirconium (I) perchlorate?

The oxidation state of zirconium is +1 since it is stated that we have zirconium(I). Converserly, perchlorate has a -1 charge which further proves that zirconium has a +1 oxidation state

The product would have a strontium diperchlorate because we know from earlier that Sr has a +2 oxidation state, and we would also have ZrOH because Zr has a +1 oxidation state and $-_{OH}$ has a -1 charge

The unbalanced equation with all chemicals formulas would be

$SrOH_2_(s) + ZrClO^{-} _4(aq) ==> Sr(ClO^{-} _4)_2(aq) + ZrOH(aq)$

Using the law of conservation of mass, it is clear that there is 1 molecule of $ClO^{-} _4$ on reactant side as compared to 2 molecules of $ClO^{-} _4$ on product side. On the other hand, the are 2 molecules of $-_{OH}$ on reactant side and 1 molecule of  $-_{OH}$ on product side. To balance we add a coefficient of 2 on $ZrClO_4$ and a coefficient of 2 on ZrOH,

$SrOH_2_(s) + 2ZrClO_4_(aq) ==> Sr(ClO_4)_2_(aq) + 2ZrOH(aq)$.