Balancing Chemical Equations

Balancing Chemical Equations

• Balancing chemical reactions involves ensuring the number of atoms for each element is equal on both sides of the equation.
• For combustion reactions, balance carbon atoms first, then hydrogen atoms, and lastly oxygen atoms.
• The goal is to adjust the coefficients in front of each molecule to achieve atomic balance.

Combustion Reaction Example: Propane (C3H8) + Oxygen (O2)

• Propane (C3H8) reacts with oxygen gas to produce carbon dioxide (CO2) and water (H2O).
• Start by balancing carbon: C3H8 requires 3 CO2 to balance the carbon atoms.
• Next, balance hydrogen: C3H8 has 8 hydrogen atoms, necessitating 4 H2O molecules.
• Balance oxygen last: 3 CO2 and 4 H2O have a combined 10 oxygen atoms, requiring 5 O2 molecules.
• The balanced equation is: C3H8 + 5 O2 -> 3 CO2 + 4 H2O.

Second Example: Butane (C4H10) + Oxygen (O2)

• Butane (C4H10) reacts with O2 to produce CO2 and H2O.
• Balance carbon: C4H10 requires 4 CO2.
• Balance hydrogen: C4H10 has 10 hydrogen atoms, therefore 5 H2O molecules are needed.
• Determine oxygen count: 4 CO2 and 5 H2O have a combined 13 oxygen atoms.
• To balance the oxygen, we can use a fraction: C4H10 + 13/2 O2 -> 4 CO2 + 5 H2O
• Multiply the whole equation by 2 to remove the fraction: 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
• The balanced equation is: 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O.

Aluminum (Al) + Hydrochloric Acid (HCl)

• Aluminum reacts with hydrochloric acid to produce aluminum chloride (AlCl3) and hydrogen gas (H2).
• Start by balancing chlorine (Cl), in this case there are three chlorine atoms on the right side, so you may be incilined to add 3HCl to the left
• The issue here is we now have an add number of hydrogens on the left (3) and an even number on the right (2), in order to address this, we must multiply our coefficents by 2
• Instead of putting three in front of HC, we put six.
• Now there are six Chorine atoms, so a two must be put in front of AlCl3
• After this, put a three in front of H2 so that there are six hydrogen atoms on both sides
• Finally, place a two in front of Al since there are 2 aluminum atomns on the right side

Gallium (Ga) + Copper Bromide (CuBr2)

• The lest commond multiple of three and two is 6, start here
• Place a two in front of GaBr3
• Place a three in front of CuBr2
• Next, place a 3 in front of Cu so there are three copper atoms on both sides
• Since there are now two gallium atoms on the right side, place a two in front of GA

Iodine (I2) + Fluorine (F2)

• Reactants are Iodine and Fluorine, product is IF7
• You want 14 fluorine atoms on either side
• Place a two in front of IF7
• Place a seven in front of F2

Sulfur Dioxide (SO2) + Oxygen gas (O2)

• Reactants are Sulfur Dioxide and Oxygen Gas, product is Sulfur Trioxide
• Four oxygen atoms on the left and three on the right
• Consider .5O2 to resolve even-odd oxygen issue, but the equation must remain with whole numbers.
• We have 2 SO2 +.5O2 -> 2SO3
• Multiply the equation to rid of the decimals
• Final balanced equation 2 SO2 + O2 -> 2 SO3

Sodium metal (Na) + Elemental Sulfur (S8)

• To balance assign a coefficient of 8 to Na2S to balance sulfur atoms
• This means we need assign a coefficient of 16 to Sodium or Na

Balancing Double Replacement Reactions

Sodium Phosphate (Na3PO4) + Magnesium Chloride (MgCl2)

• Products are Sodium Chloride and Magnesium Phosphate
• Rather than balancing each element individually, balance phosphate units as a whole.
• In this case, place a two in front of Na3Po4
• Next, place a 6 in front of NaCl to balance the equation

Potassium Sulfate (K2SO4) + Aluminum Chloride (AlCl3)

• Products are Potassium Chloride and Aluminum Sulfate
• Similar to the previous question, treat sulfate as a single unit
• We must begin by putting a 3 in front of K2SO4 to balance with our sulfate
• Next, put a oefficient of 2 in front of Aluminum Chloride to balance aluminums
• Finally, solve the problem by putting a 6 in front of KCL

Ammonia (NH3) + Oxygen gas (O2)

• This is a relatively tricky problem that requires you to double back and re-balance due to Oxygen existing on both sides of the product
• It seems like our nitrogen atoms are balanced from the outset, so continue by addressing the hydrogen (save oxygen for last). Odd left, even right, so we must double the hydrogen.
• Use a 2, in front of NH3, this means you need three H2O molecules so that there are 6 hydrogen molecules
• We must put a 2 in front of NO because there are two nitrogen atoms on the left
• Adding things up, what do you do so we can balance the oxygens -- remember it is 5 oxygens and we divide by 2, the coefficent in front of O2 (5\2)
• Multiply everything by 2, so that we can cancel 2/5 and not have fractions: 4NH3 + 5O2 => 4NO + 6H2O

Ethanol (C2H5OH) + Oxygen gas (O2)

• Products of this combustion reaction are Carbon Dioxide in Water
• Start by balancing Carbon, place a 2 in front of CO2
• A single ethanol molecule has 6 hydrogens, so place a 3 in front of H2O
• Next, factor in the oxygen from ethanol, and the goal must be 7 oxygen atoms. Solve for X when 1 + X = 7, or X = 6
• In the final step, 6/2 is 3, put a 3 in front of O2, and everything is balanced