General Chemistry

To work through a stoichiometry worksheet that focuses on limiting reagents and percent yield calculations, we will need a specific chemical reaction to analyze. For the purpose of this example, let's consider the reaction between hydrogen gas (H₂) and oxygen gas (O₂) to form water (H₂O): \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \] Let's assume we start with 5.00 grams of hydrogen gas and 32.00 grams of oxygen gas. We want to find out which reactant is the limiting reagent and calculate the theoretical yield and percent yield of water if the actual yield is 36.00 grams. **Step 1: Convert grams to moles for each reactant.** First, we need to convert the mass of each reactant to moles using their molar masses. The molar mass of hydrogen gas (H₂) is approximately 2.02 g/mol, and the molar mass of oxygen gas (O₂) is approximately 32.00 g/mol. For hydrogen: \[ \text{moles of } H_2 = \frac{5.00 \text{ g}}{2.02 \text{ g/mol}} \approx 2.475 \text{ moles of } H_2 \] For oxygen: \[ \text{moles of } O_2 = \frac{32.00 \text{ g}}{32.00 \text{ g/mol}} = 1.000 \text{ moles of } O_2 \] **Step 2: Determine the limiting reagent.** Using the balanced chemical equation, we can see that it takes 2 moles of hydrogen gas to react with 1 mole of oxygen gas. We need to compare the mole ratio of the reactants we have to the mole ratio required by the balanced equation. For hydrogen, we have 2.475 moles, and for oxygen, we have 1.000 mole. The ratio required by the equation is 2:1, so we divide the moles of hydrogen by 2 and compare it to the moles of oxygen: \[ \frac{2.475 \text{ moles of } H_2}{2} = 1.2375 \text{ moles of } H_2 \] Since 1.2375 moles of hydrogen is more than the 1.000 mole of oxygen we have, oxygen is the limiting reagent. **Step 3: Calculate the theoretical yield of water.** Using the limiting reagent (oxygen), we can calculate the theoretical yield of water. According to the balanced equation, 1 mole of oxygen gas produces 2 moles of water. \[ \text{Theoretical yield of } H_2O = 1.000 \text{ mole of } O_2 \times \frac{2 \text{ moles of } H_2O}{1 \text{ mole of } O_2} \] \[ \text{Theoretical yield of } H_2O = 2.000 \text{ moles of } H_2O \] Now, convert moles of water to grams using the molar mass of water (approximately 18.02 g/mol): \[ \text{Theoretical yield of } H_2O = 2.000 \text{ moles of } H_2O \times 18.02 \text{ g/mol} \] \[ \text{Theoretical yield of } H_2O = 36.04 \text{ g} \] **Step 4: Calculate the percent yield.** The percent yield is calculated using the actual yield and the theoretical yield: \[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% \] Given that the actual yield is 36.00 grams: \[ \text{Percent yield} = \left( \frac{36.00 \text{ g}}{36.04 \text{ g}} \right) \times 100\% \] \[ \text{Percent yield} \approx 99.89\% \] In conclusion, oxygen is the limiting reagent, the theoretical yield of water is 36.04 grams, and the percent yield of the reaction is approximately 99.89%.

Question

Work through stoichiometry worksheet 2 answer key, focusing on limiting reagents and percent yield c...

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