Video Transcript
The dark green solid molybdenum(V) chloride MoCl5 reacts with metallic gray molybdenum to produce the orange crystals of molybdenum(II) chloride Mo6Cl12 according to the following equation: 𝑥MoCl5 plus 𝑦Mo react to produce 5Mo6Cl12. The coefficients 𝑥 and 𝑦 are both whole numbers. Find the value of 𝑥. Find the value of 𝑦.
We need to determine what value of 𝑥 and 𝑦 will balance this chemical equation. Let’s begin this problem by identifying the elements involved in the reaction. Every element has a chemical symbol. And every chemical symbol is either a capital letter or a capital letter followed by a lowercase letter. What we need to do is look at the chemical equation and break it down into the individual letter groups.
Starting at the beginning of the equation, we see capital M followed by lowercase o. This is a chemical symbol. This chemical symbol appears two more times in the equation. We also see capital C lowercase l, which appears twice in the equation. We can make a list of these elements using the periodic table to match the chemical symbols we identified with the element name. Mo is molybdenum, and Cl is chlorine.
Next, we will count the number of atoms of each element on both sides of the reaction. We can make a chart to help us organize this information. The chart consists of our element list down the left-hand side and the chemical reaction written across the top. We can divide the chart at the arrow separating the reaction into reactants and products.
Now we can begin counting atoms. On the reactants side, we can see one molybdenum atom in the first molecule and one molybdenum atom in its pure form. On the products side, there are six molybdenum atoms in molybdenum(II) chloride. However, there are five molecules of molybdenum(II) chloride. If one molecule contains six atoms, then five molecules will contain 30 total atoms.
Now let’s take a look at chlorine. On the reactants side, there are five atoms of chlorine in the first molecule. On the products side, here are 12 atoms of chlorine in each molybdenum(II) chloride molecule. But once again, we are dealing with five molecules of molybdenum(II) chloride. If one molecule contains 12 atoms of chlorine, five molecules will contain 60 total atoms of chlorine.
Now that we have counted the number of atoms of each element on both sides of the reaction, we are ready to add coefficients in front of the molecules on the reactants side in order to balance the reaction. A reaction is balanced when the number of atoms of each element are the same on both sides of the reaction. We need to decide which element to begin with when balancing this reaction. One helpful tip is to look for elements which exist in their pure form in the reaction. These elements should be balanced last.
We should balance these elements last because placing a coefficient in front of the pure form in the chemical equation will only affect the number of atoms of that element. Any other elements in the chemical equation will remain balanced. This means that we should begin by balancing the chlorine atoms. We can place a coefficient of 12 in front of the molybdenum(V) chloride. This gives us 60 chlorine atoms on our reactants side. With 60 chlorine atoms on both sides of the reaction, the chlorine is balanced.
Placing a coefficient of 12 in front of molybdenum(V) chloride also affected the number of molybdenum atoms, meaning that we now have 12 molybdenum atoms from the molybdenum(V) chloride molecules. We now need to determine what coefficient value in front of the molybdenum metal would make the molybdenum atoms balanced.
We can set the molybdenum atoms on the reactants side equal to the molybdenum atoms on the products side and rearrange to solve for 𝑦. In doing so, we determine that 𝑦 must have a value of 18. With a coefficient of 18, there are now a total of 30 atoms of molybdenum on the reactants side. This means that the molybdenum atoms are balanced.
Upon balancing the chemical equation, we have determined that the coefficients 𝑥 and 𝑦 have the values 𝑥 equals 12 and 𝑦 equals 18.
