How to Calculate Theoretical Yield Step by Step

Welcome to our lesson on balanced equations and limiting reagents!Let's start with a simple reaction between hydrogen and oxygen.When balancing equations, we follow three important rules.First, we count the atoms on each side. Then we add coefficients to balance them, never changing the subscripts.Now let's look at a practical example to understand limiting reagents.We have 4.0 grams of hydrogen and 32.0 grams of oxygen.To compare these reactants, we need to convert them to moles using their molar masses.Let's perform the conversion calculations.Now we can compare the actual mole ratio to the required ratio from our balanced equation.Since we need a two-to-one ratio of hydrogen to oxygen, and we have slightly less than that, oxygen is not our limiting reagent - hydrogen is.Now that we've identified our limiting reagent, we can move on to stoichiometric calculations.Now that we have our balanced equation and limiting reagent, we can perform our stoichiometric calculations.From our balanced equation, we can establish the mole ratios between reactants and products.Let's work with our given information. We have 4.0 grams of hydrogen gas as our limiting reagent.First, we need to convert the mass of hydrogen to moles using its molar mass.Next, we use the mole ratio from our balanced equation to determine the moles of water product. Since the ratio of hydrogen to water is two to two, or one to one, we'll get the same number of moles.Throughout these calculations, we must maintain proper significant figures. Our initial measurement of 4.0 grams has two significant figures, so our final answer should also have two significant figures.Before moving on, let's verify our calculations by checking our units, significant figures, and proper application of mole ratios.Now that we have our theoretical number of moles, we can proceed to calculate the final mass of our product.Now that we have our theoretical moles, let's convert to the final mass.To convert moles to grams, we multiply by the molar mass.Let's perform our calculation. Zero point seven five moles times eighteen point zero two grams per mole.Following significant figure rules, we round our answer to thirteen point five two grams.Let's verify our answer using these important checks.For significant figures, we follow the rule of limiting our answer to the least number of significant figures in our given values.Finally, let's verify our units cancel correctly.