Video Transcript
Calcium oxide can be produced by the thermal decomposition of calcium carbonate according to the equation below. CaCO3 reacts to form CaO plus CO2. What is the minimum amount of calcium carbonate needed to produce 1.12 grams of calcium oxide? The molar mass of calcium is 40 grams per mole, carbon is 12 grams per mole, and oxygen is 16 grams per mole.
In this question, we are being asked to determine the amount in grams of one substance in a reaction when given the amount in grams of a second substance in the same reaction. To solve this problem, we will need to figure out the relationship between calcium carbonate and calcium oxide by using the provided chemical equation. We should first make sure that the chemical equation is balanced. Let’s count up how many of each type of atom we have on either side of the equation.
On the reactant side, we have one calcium atom. And on the product side, we also have one calcium atom. On the reactant side, we have one carbon atom. And on the product side, we also have one carbon atom. Finally, on the reactant side, we have three oxygen atoms. And on the product side, we have one oxygen atom in CaO plus two oxygen atoms in CO2 for a total of three oxygen atoms on the product side. So we have confirmed that this chemical equation is already balanced.
We know that 1.12 grams of calcium oxide is produced. The chemical formula of calcium oxide is CaO. So let’s write this mass below it. To solve this problem, we must determine the amount of calcium carbonate needed. The chemical formula of calcium carbonate is CaCO3. There are three main steps in our problem solving process. First, we need to convert 1.12 grams of calcium oxide to moles. Next, we’ll use the molar ratio to find the number of moles of CaCO3 that reacted. And then we will convert the number of moles of CaCO3 to grams. To convert 1.12 grams of CaO to moles, we need to use the following equation. In this equation, 𝑛 is the number of moles, lowercase 𝑚 is the mass in grams, and uppercase 𝑀 is the molar mass in grams per mole.
The molar mass of calcium oxide can be calculated by adding together the average molar masses of the atoms that make up one formula unit. A formula unit of calcium oxide contains one calcium atom, which has an average molar mass of 40 grams per mole, and one oxygen atom, which has an average molar mass of 16 grams per mole. Combining these masses gives us a molar mass of 56 grams per mole for calcium oxide. Let’s substitute 1.12 grams of CaO and the molar mass 56 grams per mole into the equation and divide. The result is 0.02 moles of CaO. In a balanced chemical equation, the stoichiometric coefficients placed in front of chemical formulas represent the number of moles of each species needed for a complete reaction.
In the provided equation, there are no coefficients, which means that there is one mole of each substance for a complete reaction. Therefore, one mole of calcium carbonate is required to produce one mole of calcium oxide. Let’s write this molar ratio as a fraction so that we can use it as a conversion factor. We should write one mole of CaCO3 in the numerator because the desired units in step (2) are moles of CaCO3. Now we can take our answer from step (1), 0.02 moles of CaO, and multiply by the molar ratio. The units moles of CaO cancel. And the answer is 0.02 moles of CaCO3. Now we’re ready to complete the last step, which is to convert our answer into grams of CaCO3.
To find the amount in grams of CaCO3, we’ll need to take the number of moles and multiply by the molar mass of CaCO3. One formula unit of calcium carbonate contains one calcium atom, one carbon atom, and three oxygen atoms. One calcium atom has an average molar mass of 40 grams per mole, and one carbon atom has an average molar mass of 12 grams per mole. Each oxygen atom has an average molar mass of 16 grams per mole, so the combined mass of three oxygen atoms is 48 grams per mole. After adding these masses together, we find that the molar mass of calcium carbonate is 100 grams per mole. Now we can substitute the number of moles of CaCO3 and the molar mass of CaCO3 into the equation. After multiplying, the result is two grams of CaCO3.
What is the minimum amount of calcium carbonate needed to produce 1.12 grams of calcium oxide? The answer is two grams.
