Question #45791

1 Answer
Jan 5, 2017

Answer:

Here's what I got.

Explanation:

Start by writing down the unbalanced chemical equation

#"NH"_ (3(g)) -> "N"_ (2(g)) + "H"_ (2(g))#

Your goal when balancing a chemical equation is to make sure that all the atoms that are present on the reactants' side, i.e. on the left side of the reaction arrow, are also present on the products' side, i.e. on the right side of the arrow.

Notice that at this point, the reactants' side contains

  • one atom of nitrogen, #1 xx "N"#
  • three atoms of hydrogen, #3 xx "H"#

The products' side contains

  • two atoms of nitrogen, # 2 xx "N"#
  • two atoms of hydrogen, #2 xx "H"#

Notice that you can balance the atoms of nitrogen by multiplying the ammonia molecule by #color(red)(2)#. This will give you

#color(red)(2)"NH"_ (3(g)) -> "N"_ (2(g)) + "H"_ (2(g))#

The reactants' side contains

  • two atoms of nitrogen, #color(red)(2) xx (1 xx "N") = 2 xx "N"#
  • six atoms of hydrogen, #color(red)(2) xx (3 xx "H") = 6 xx "H"#

The products' side contains

  • two atoms of nitrogen, # 2 xx "N"#
  • two atoms of hydrogen, #2 xx "H"#

Now balance the hydrogen atoms by multiply the hydrogen molecule by #color(blue)(3)#. This will give you

#color(red)(2)"NH"_ (3(g)) -> "N"_ (2(g)) + color(blue)(3)"H"_ (2(g))#

The reactants' side contains

  • two atoms of nitrogen, #2 xx "N"#
  • six atoms of hydrogen, #6 xx "H"#

The products' side contains

  • two atoms of nitrogen, # 2 xx "N"#
  • two atoms of hydrogen, #color(blue)(3) xx (2 xx "H") = 6 xx "H"#

You can now say that the chemical equation is balanced

#color(darkgreen)(ul(color(black)(2"NH"_ (3(g)) -> 2"N"_ (2(g)) + 3"H"_ (2(g)))))#

Finally, does this reaction represent a decomposition reaction?

As you know, a decomposition reaction is said to take place when a single compound breaks down into two or more elements or simpler compounds.

https://www.emaze.com/@ALTROZRZ/DECOMPOSITION-REACTION.pptx

In your case, ammonia, #"NH"_3#, breaks down into its constituent elements, nitrogen gas, #"N"_2#, and hydrogen gas, #"H"_2#.

Therefore, you can say that the given reaction is indeed a decomposition reaction.