Balance the carbon - C₄H₁₀ + O₂ → 4CO₂ + H₂O

Balance the hydrogen - C₄H₁₀ + O₂ → 4CO₂ + 5H₂O

Now balace the oxygen - C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O

fractions are not allowed in the correctly balanced equation so multiply through by 2 to remove the fraction.

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Response B: Coefficient = 13

First balance the carbon

_C₂H₂(g) + _O₂(g) → 2CO₂(g) + _H₂O(g)

Then the hydrogen

_C₂H₂(g) + _O₂(g) → 2CO₂(g) + 1H₂O(g)

However there are now an odd number of oxygen atoms. This will not make a whole number of O₂ molecules. Therefore the whole equation must be multiplied through by 2.

2C₂H₂(g) + _O₂(g) → 4CO₂(g) + 2H₂O(g)

Now balance the oxygen atoms

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

Response D: 5 x O₂

Balance the sulfur

2H₂S + _O₂ → 2SO₂ +_H₂O

Then balance the hydrogen

2H₂S + _O₂ → 2SO₂ + 2H₂O

Now work backwards and balance the oxygen

2H₂S + 3O₂ → 2SO₂ + 2H₂O

Response C: 3 moles of oxygen

Assume w = 1, balance the carbon atoms:

C₄H₉OH + xO₂ → 4CO₂ + zH₂O

balance the hydrogen atoms

C₄H₉OH + xO₂ → 4CO₂ + 5H₂O

Balance the oxygen atoms, do not forget the oxygen atom in the butanol:

C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O

Response A : 6

Inspection of the equation shows that the Pb and the S are balanced, but the oxygen atoms are unbalanced.

The right hand side must provide an even number of oxygen atoms while maintaining the coefficients of PbO(s) and SO₂(g) equal.

Double the right hand side:

PbS(s) + O₂(g) → 2PbO(s) + 2SO₂(g)

Now balance the PbS(s) and the O₂(g)

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

Response B : sum of coefficients = 9

First write down the formulae for each component of the reaction

NO + O₂ → NO₂

Now inspect the number of atoms on either side. There are three oxygen atoms on the left and only two on the right. Multiply the NO by 2 and the NO₂ by 2

2NO + O₂ → 2NO₂

The equation is balanced!

From the equation: Ca₃(PO₄)₂ + SiO₂ → CaSiO₃ + P₄O₁₀

It may be seen that all of the phosphorus atoms from the calcium phosphate end up in the phosphorus pentoxide.

Therefore the coefficient for Ca₃(PO₄)₂ must be 2 and working forwards the coefficient for CaSiO₃ must be 6 to equalise the calcium atoms on both sides.

2Ca₃(PO₄)₂ + SiO₂ → 6CaSiO₃ + P₄O₁₀

Working backwards then the coefficient for the SiO₂ must also be = 6

2Ca₃(PO₄)₂ + 6SiO₂ → 6CaSiO₃ + P₄O₁₀

From the equation: C₆H₆ + O₂ → CO₂ + H₂O

6 carbon atoms in the benzene must make 6 carbon dioxide molecules

C₆H₆ + O₂ → 6CO₂ + H₂O

And 6 hydrogen atoms from the benzene must produce 3 water molecules

C₆H₆ + O₂ → 6CO₂ + 3H₂O

Counting back the oxygens on the right hand side = 15. This would make an odd number of O₂ molecules so the whole eqation must be mutiplied by 2

2C₆H₆ + O₂ → 12CO₂ + 6H₂O

Now there are 30 oxygen atoms on the right hand side = 15 x O₂ molecules, giving

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Therefore the coefficient for oxygen = 15

By inspection of the equation: KClO₃ → KClO₄ + KCl

It may be seen that the oxygen atoms must be equalised. They only occur in two compounds therefore mutiplying each by its opposite number of oxygens makes them equal, i.e mutiply KClO₃ by 4, and KClO₄ by 3.

4KClO₃ → 3KClO₄ + KCl

The equation is now balanced so the coefficient for potassium chlorate (V) = 4

Inspection of the equation: Sb₂S₃ + O₂ → SO₂ + Sb₂O₃

Shows us that all of the sulfur in the antimony sulfide Sb₂S₃ turns to sulfur dioxide SO₂

Sb₂S₃ + O₂ → 3SO₂ + Sb₂O₃

There are now 9 x oxgen atoms on the right hand side - an odd number so the equation must be multiplied by 2

2Sb₂S₃ + O₂ → 6SO₂ + 2Sb₂O₃

now there are 18 oxygen atoms on the right hand side = 9 x O₂ molecules

2Sb₂S₃ + 9O₂ → 6SO₂ + 2Sb₂O₃

Therefore the coefficient for oxygen = 9

Step 2: write down the symbols for each component of the equation

Na + O₂ → Na₂O

Step 3: balance the numbers of atoms on either side of the arrow

4Na + O₂ → 2Na₂O

Step 2: write down the symbols for each component of the equation

Na₂CO₃ + HCl → NaCl + CO₂ + H₂O

Step 3: balance the numbers of atoms on either side of the arrow

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

Step 2: write down the symbols for each component of the equation

Ca(NO₃)₂ → CaO + NO₂ + O₂

Step 3: balance the numbers of atoms on either side of the arrow

2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂

Step 2: write down the symbols for each component of the equation

Ca(OH)₂ + HCl → CaCl₂ + H₂O

Step 3: balance the numbers of atoms on either side of the arrow

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Step 2: write down the symbols for each component of the equation

Fe + Cl₂ → FeCl₃

Step 3: balance the numbers of atoms on either side of the arrow

2Fe + 3Cl₂ → 2FeCl₃

Word equation:

sulfuric acid and sodium hydroxide → sodium sulfate + water

formulae:

NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)

Balanced:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Here we can see that the sodium ions (Na⁺) in the sodium hydroxide solution remain as sodium ions in the sodium sulfate solution - they are SPECTATOR IONS.

The same applies to the sulfate ions SO₄^2-. They start off as sulfate ions in the sulfuric acid solution and end up as sulfate ions in the sodium sulfate solution - they are also spectator ions. The only ions that actually react together are the hydrogen ions from the sulfuric acid and the hydroxide ions from the sodium hydroxide:

H⁺(aq) + OH^-(aq) → H₂O(l)

Word equation:

silver nitrate + sodium chloride → sodium nitrate + silver chloride

formulae:

AgNO₃(aq) + NaCl(aq) → NaNO₃(aq) + AgCl(s)

Already balanced:

Here we can see that the silver chloride is formed as a solid precipitate. These ions must then be removed from the solution. All of the other ions remain behind as spectator ions in solution

Ag⁺(aq) + Cl^-(aq) → AgCl(s)

Barium chloride and sodium sulfate react together to form a precipitate of barium sulfate - this is used to identify the sulfate group.

The unbalanced equation

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + NaCl(aq)

balanced

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

BUT, the sodium chloride is still in solution so its ions have not reacted.

The only ions involved in reaction are those that come together to make the precipitate:

Ba²⁺(aq) + SO₃₄^-(aq) → BaSO₄(s)

iron(III) sulfate and sodium hydroxide react together to form a precipitate of iron(III) hydroxide - this is used to identify the iron(III) ion.

The unbalanced equation

NaOH(aq) + Fe₂(SO₄)₃(aq) → Na₂SO₄(aq) + Fe(OH)₃(s)

balanced

6NaOH(aq) + Fe₂(SO₄)₃(aq) → 3Na₂SO₄(aq) + 2Fe(OH)₃(s)

BUT, the sodium ions have not reacted and neither have the sulfate ions - the sodium sulfate 'produced' is in solution.

The only ions involved in reaction are those that come together to make the precipitate:

Fe³⁺(aq) + 3OH^-(aq) → Fe(OH)₃(s)

This is the reaction used to prepare chlorine in the laboratory:

manganese(IV) oxide and hydrochloric acid → manganese(II) chloride + chlorine + water

The balanced equation:

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + 2H₂O(l)

Two of the chloride ions from the left hand side are spectator ions whereas all of the other ions are involved in reaction:

MnO₂(s) + 4H⁺(aq) + 2Cl^-(aq) → Mn²⁺(aq) + Cl₂(g) + 2H₂O(l)

If you check out the charge totals on both sides it may be seen that they are equal.