Question #2d887

2 Answers
Feb 27, 2017

Answer:

#2K(s) + 2H_2O(l) rarr 2KOH (aq) + H_2(g)#

Explanation:

Or, if you need an ionic equation:

#2K(s) + 2H_2O(l) rarr 2K^+(aq) + 2OH^-## (aq) + H_2(g)#

Feb 27, 2017

Answer:

#2"K"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"KOH"_ ((aq)) + "H"_ (2(g)) uarr#

Explanation:

You know that this reaction has solid potassium, #"K"#, and liquid water, #"H"_2"O"#, as reactants and aqueous potassium hydroxide, #"KOH"#, and hydrogen gas, #"H"_2#, as products.

The first thing to do here is to write the unbalanced chemical equation -- remember, reactants are added to the left of the reaction arrow and products are added to the right of the reaction arrow.

#"K"_ ((s)) + "H"_ 2"O"_ ((l)) -> "KOH"_ ((aq)) + "H"_ (2(g)) uarr#

Notice that each chemical species that takes part in the reaction has a symbol added to describe its state

  • solid potassium #-> (s)#
  • liquid water #-> (l)#
  • aqueous potassium hydroxide #-> (aq)#
  • gaseous hydrogen #-> (g)#

Now, in order to balance this chemical equation, add a coefficient of #2# for potassium, water, and potassium hydroxide

#2"K"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"KOH"_ ((aq)) + "H"_ (2(g)) uarr#

This will ensure that all the atoms that are present on the left side of the reaction arrow will also be present on the right side of the reaction arrow.

More specifically, you will have

  • #overbrace(2 xx "K")^(color(purple)("from 2K")) -> overbrace(2 xx "K")^(color(purple)("from 2KOH"))" " implies# potassium is balanced
  • #overbrace(2 xx 2 xx "H")^(color(blue)("from 2H"_2"O")) -> overbrace(2 xx "H")^(color(blue)("from 2KOH")) + overbrace(2 xx "H")^(color(blue)("from H"_2))" "implies# hydrogen is balanced
  • #overbrace(2 xx "O")^(color(darkgreen)("from 2H"_2"O")) -> overbrace(2 xx "O")^(color(darkgreen)("from 2KOH"))" " implies# oxygen is balanced