Reactions - AP Chemistry

Balancing reactions is best acheived by using a stepwise approach. It is useful to work through each atom making sure it is present in a balanced fashion on both sides of the equation. Starting with Fe, Fe2O3 is the only molecule with Fe present on the left side of the equation and FeCl3 is the only molecule with Fe present on the right side of the equation. Thus the molar ratio of Fe2O3 to FeCl3 must be 1:2. Moving on to O, the O on the left side of the equation exists as Fe2O3 and H2O on the right. The molar ratio of Fe2O3 to H2O is therefore 1:3. Cl exists in HCl on the left and FeCl3 on the right. Thus the molar ratio of HCl to FeCl3 must be 3:1. To ensure that these molar ratios are maintaned the balanced formula is then determined to be Fe2O3 + 6HCl -> 2FeCl3 + 3H2O

The first thing to do is to balance the Cl (3 on the left; one on the right) ← add 3 for NH4Cl

Now, there are 3 N on the right and only one on the left; add a 3 to the NOH5

Check to see that the Fe, H, and O balance, which they do

The reaction balances out to 2C2H6(s) + 7O2(g) → 6H2O(l) + 4CO2(g). For every 2 moles of C2H6 you can produce 6 moles of H2O. Giving you a total production of 12 moles when you have 4 moles of C2H6.

Begin by balancing the equation. There are many ways to do this, but one method that is particularly useful is to assume that you have 1 mole of your hydrocarbon (propane), and balance the equation from there. It may be necessary to manipulate an equation further if you end up with fractions, but all you will need to do is multiply by an integer if that is the case.

First, we will balance the carbons. There are three carbons in propane, so we will make sure there are three carbons on the right side of the arrow as well:

This is not complete. We will next balance the hydrogens. There are eight hydrogens on the left side of the equation, so:

The last step is to balance the oxygens on the left and right side of the equation

Our equation is balanced and all coefficients are integers. If we begin with one mole of propane, we will produce four moles of water.

Potassium is a Group I element, so to get to a filled valence shell, it will lost one electron, yielding .

Arsenic is a Group 5 element, so it needs to gain three electrons to obtain a filled valence shell, yielding .

In order to balance out the charges, the resultant salt will be .

First, we will balance the equation:

Since chlorine is in excess, we know that the limiting reagent is iron.

First, we must know what ferrous sulfate is. Ferrous refers to , and sulfate has the formula . When we combine the two together we get .

Calcium is a divatent cation and iodide is a monovalent anion, so their salt is . The ion exchange reaction is then:

When balancing equations, the goal is to make sure that the same atoms, in both type and amount, are on both the reactant and product side of the equation. A helpful approach is to write down the number of atoms already on both sides of the unbalanced equation. This way, you can predict which compounds need to be increased on which side in order to balance the equation. It also helps to balance oxygen and hydrogen last in the equation.

In this reaction, we can balance as follows.

Reactants: 1K, 1Mn, 1Cl, 4O, 1H

Products: 1K, 1Mn, 5Cl, 1O, 2H

So, we will need to increase H2O and HCl. The final balanced equation is written below.

Iron (II) has a positive two charge: .

Phosphate has a negative three charge: .

The initial compound must be constructed to cancel these charges. The dissociation is: .

To balance an equation, we need to make sure there is the same amount of elements to the left of the arrow as there is to the right. We also need all the charges to balance out. We notice right away that there are three chlorine atoms on the left, but only one on the right.

(1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

(1Na, 1O, 1H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 3Cl)

This causes us to have an imbalance of sodium, which we can correct by manipulating NaOH.

(3Na, 3O, 3H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 1Cl)

This is the final balanced equation. Note that it is usually easiest to manipulate oxygen and hydrogen last, since they are often involved in multiple molecules.

Calcium is in the second group of the periodic table, and is therefore going to have a oxidation number. Hydroxide ions have a charge. Calcium hydroxide will have the formula .

Chloride ions have a charge and hydrogen ions have a charge. The formula for hydrochloric acid is .

On the products side, water has the formula and calcium chloride has the formula .

Now that we know all of the formulas, we can write our reaction:

In order to balance the chloride atoms, we need to add coefficients.

The balanced reaction for the combustion of pentane is:

When balanced, oxygen gas has a coefficient of eight.

To balance the equation, it is easiest to leave oxygen and hydrogen for last. This means we should start with carbon.

Now that carbon is balanced, we can look at hydrogen.

Finally, we can balance the oxygen.

The final reaction uses five carbon atoms, twelve hydrogen atoms, and sixteen oxygen atoms per side.

The net ionic equation is derived by removing all spectator ions from the total ionic equation (in which all ions are listed). To put it another way, the net ionic equation involves only the ions that participate in a reaction which, in this case, is the precipitation of barium sulfate.

Begin by writing all aqueous compounds in their dissociated (ionic) forms.

Cancel out any ions that appear in equal quantities on both sides of the equation. In this case, we can cancel the nitrate and potassium ions.

This is our net ionic equation.

Combustion is the chemical reaction of a hydrocarbon with molecular oxygen, and it always produces carbon dioxide and water. Knowing the reactants and products, the unbalanced equation must be:

We start by balancing the hydrogens. Since there are 10 on the left and only 2 on the right, we put a coefficient of 5 on water.

Similarly, we balance carbons by putting a 4 on the carbon dioxide.

To find the number of oxygens on the right, we multiply the 4 coefficient by the 2 subscript on O (which gets us 8 oxygens) and then add the 5 oxygens from the 5 water molecules to get a total of 13. The needed coefficient for on the left would then have to be 13/2.

Because fractional coefficients are not allowed, we mutiply every coefficient by 2 to find our final reaction:

When we check the activity series, it is fairly easy to see that aluminum metal is more reactive than zinc metal. So, in this case, the two metals undergo a redox reaction, where the aqueous is reduced to solid , and the solid is oxidized to aqueous . These charges are the common oxidation states for zinc and aluminum and should be memorized.

Because is the new species, it bonds with 3 ions. The unbalanced equation is:

We note that there are 2 chlorine atoms on the left and 3 chlorine atoms on the right. To balance, we use a 3 coefficient on the left and a 2 coefficient on the right. This gives a total of 6 chlorine atoms on eahc side.

However, now we have also increased the amounts of zinc and aluminum. We copy the necessary coefficients to balance those—2 for aluminum on the left, 3 for zinc on the right—and we are done:

The balanced chemical equation is:

The mole ratio of iron oxide to solid iron is 1:2. You can set up the following proportion to solve:

First, balance the carbon atoms in the reactants and products. Next, balance the hydrogen atoms. Third, balnace the oxygen atoms. Since stoichiometric coefficients are written as integers, double everything to remove the decimal/fraction.

One molecule of , balances with molecules and molecules. On the right side of the equation, there are a total of 18 oxygen atoms, which equates to molecules on the left side of the equation.

As when balancing the vast majority of chemical reaction, leave oxygen as the last element to balance. Start by balancing the hydrogen atoms - this requires a 4 in front of water on the products side. Next, balance the three carbon atoms on the reactant side by adding a coefficient of 3 to the carbon dioxide in the products. Finally balance the oxygen atoms by placing a 5 in front of the oxygen gas on the reactant side. Double check to make sure there is a 1:1 ratio of each element and that all coefficients are whole numbers. Thus the balanced chemical reaction is:

Balance aluminum by adding a coefficient of 2 to the left side:

Balance iron by adding a coefficient of 2 to the right side:

The equation is balanced.