27. Chemical Equations: Balancing

Question

1) When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:

carbon dioxide (g) + potassium hydroxide (aq) potassium carbonate (aq) + water (l)

2) When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:

carbon (graphite) (s) + oxygen (g) carbon dioxide (g)

3) When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:

bromine trifluoride (g) bromine (g) + fluorine (g)

Answer 1

First you have to write out the equations using chemical formulas.

1) CO₂(g) + KOH(aq) --> K₂CO₃(aq) + H₂O(l)

Potassium has a charge of +1 and hydroxide has a charge of -1 which cancels out creating a net ionic charge of 0 so potassium hydroxide is KOH. Carbonate has a charge of -2 so to even that out you have to add 2 potassiums since potassium only has a charge of +1 which makes potassium carbonate K₂CO₃.

When balancing an equation both sides should have the same number of elements. On the right side there are 2 potassiums but on the left side there is only 1 so you can put a 2 in front of the KOH. Everything else is equal on both sides so the balanced equation is:

CO₂(g) + 2KOH(aq) --> K₂CO₃(aq) + H₂O(l)

2) C(s) + O₂(g) --> CO₂(g)

Oxygen is O₂(g) instead of O(g) because it is a diatomic element so it normally comes in 2. Other diatomic elements are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine. This equation is already balanced because it has equal numbers of carbons and oxygens on both sides of the equation.

3) BrF₃(g) --> Br₂(g) + F₂(g)

Again we have diatomic elements but this times they are bromine and fluorine so they have a 2 next to them. To balance this equation first we put a 2 in front of the BrF₃ because there are 2 bromines on the right side of the equation but only 1 on the left side. Then there are 6 fluorines on the left side so to balance that we put a 3 in front of the F₂(g) so there can be 6 fluorines on the right side as well. Everything is equal on both sides now so the balanced equation is:

2BrF₃(g) --> Br₂(g) + 3F₂(g)