How do you balance #SrBr_2 + (NH_4)_2CO_3 - SrCO_3 + NH_4Br#?

Question

How do you balance #SrBr_2 + (NH_4)_2CO_3 -> SrCO_3 + NH_4Br#?

Kai Bowman · Accepted Answer

undefined The balanced chemical equation for the reaction is:

$$ SrBr_2 + (NH_4)_2CO_3 \rightarrow SrCO_3 + 2NH_4Br $$

Serenity Allen · Answer

#"SrBr"_2 + ("NH"_4)_2"CO"_3##rarr##"SrCO"_3 + color(red)(2)"NH"_4"Br"#

Balance the chemical equation: #"SrBr"_2 + ("NH"_4)_2"CO"_3##rarr##"SrCO"_3 + "NH"_4"Br"#

The word equation is strontium bromide #("SrBr"_2")# plus ammonium carbonate #(("NH"_4)_2"CO"_3)"# produce strontium carbonate #("SrCO"_3)# plus ammonium bromide #("NH"_4"Br")#.

This is a double replacement reaction, in which the positive ions and negative ions switch partners. In this example, all of the reactants and products are ionic compounds.

A balanced equation must have the same number of elements of each compound on both sides of the equation.

When balancing equations with ionic compounds, a good start is to look for and balance polyatomic ions. They can be treated as a single quantity when balancing. This equation has two polyatomic ions; the ammonium ion #("NH"_4^(+)")#, and the carbonate ion #("CO"_3^(2-)")#.

#"SrBr"_2 + ("NH"_4)_2"CO"_3##rarr##"SrCO"_3 + "NH"_4"Br"#

There are two ammonium ions on the left side and one on the right side. We CANNOT change the formulas, but we can change the amount by adding coefficients. In the case of #"NH"_4"Br"#, we can add a coefficient of 2 in front of the compound.

#"SrBr"_2 + ("NH"_4)_2"CO"_3##rarr##"SrCO"_3 + color(red)(2)"NH"_4"Br"#

Now there is one carbonate ion on both sides of the equation, two bromide ions #("Br"^(-)")# on both sides, and one strontium ion #("Sr"^(2+)")# on both sides. So the equation is now balanced.

How do you balance #SrBr_2 + (NH_4)_2CO_3 -&gt; SrCO_3 + NH_4Br#?

How do you balance #SrBr_2 + (NH_4)_2CO_3 -> SrCO_3 + NH_4Br#?