Question #2d887
2 Answers
Explanation:
Or, if you need an ionic equation:
Explanation:
You know that this reaction has solid potassium,
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"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