# How would you interpret the given equation in terms of relative numbers of representative particles, numbers of moles, and masses of reactants and products? 2K(s) +2H_2O(l) -> 2KOH(aq) + H_2(g)

Jun 6, 2018

#### Answer:

Well, you got a redox reaction that is stoichiometrically balanced...

${\underbrace{2 K \left(s\right) + 2 {H}_{2} O \left(l\right)}}_{\text{114.2 g of reactants" rarr underbrace(2KOH(aq) +H_2(g)uarr)_"114.2 g of products}}$

#### Explanation:

${\underbrace{2 K \left(s\right) + 2 {H}_{2} O \left(l\right)}}_{\text{114.2 g of reactants" rarr underbrace(2KOH(aq) +H_2(g)uarr)_"114.2 g of products}}$

Stoichiometry, the principle that garbage in equals garbage out, insists that mass is conserved, and the equation necessarily represents this. The equation also illustrates conservation of CHARGE. Potassium metal is OXIDIZED, and water is reduced. And we could represent this by the formalism of redox reaction...

${\underbrace{K \left(s\right) \rightarrow {K}^{+} + {e}^{-}}}_{\text{the oxidation reaction with respect to the metal}}$ $\left(i\right)$

${\underbrace{{H}_{2} O \left(l\right) + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right) + H {O}^{-}}}_{\text{the reduction reaction with respect to water}}$ $\left(i i\right)$

And (i)+(ii)….gives...

$K \left(s\right) + {H}_{2} O \left(l\right) \rightarrow K O H \left(a q\right) + \frac{1}{2} {H}_{2} \left(g\right)$

I leave it to you to interpolate the NUMBER of metal, water, dihydrogen atoms, and potassium hydroxide formula units.