Sorry a lot of my problems didn’t format well but the answers is at the bottom.
To solve this problem, you need to use the concept of **limiting reagent** and **percent yield**. A limiting reagent is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed¹. The percent yield is the ratio of the actual yield (the amount of product obtained from the reaction) to the theoretical yield (the maximum amount of product that can be formed from the limiting reagent) expressed as a percentage¹.
Here are the steps to follow:
- Step 1: Convert the masses of V$_2$O$_5$ and Ca to moles using their molar masses.
- Step 2: Find the limiting reagent by comparing the mole ratios of the reactants from the balanced equation and the given masses.
- Step 3: Use the limiting reagent to calculate the theoretical yield of CaO using the mole ratio from the balanced equation and the molar mass of CaO.
- Step 4: Use the actual yield and the theoretical yield to calculate the percent yield of CaO.
- Step 5: Use the limiting reagent and the excess reagent to calculate the amount of the limiting reagent and the excess reagent that remain after the reaction.
Here are the calculations:
- Step 1: The molar masses of V$_2$O$_5$ and Ca are 181.88 g/mol and 40.08 g/mol, respectively. Therefore, the moles of V$_2$O$_5$ and Ca are:
$$\text{moles of V}_2\text{O}_5 = \frac{\text{mass of V}_2\text{O}_5}{\text{molar mass of V}_2\text{O}_5} = \frac{10.00 \text{ g}}{181.88 \text{ g/mol}} = 0.0550 \text{ mol}$$
$$\text{moles of Ca} = \frac{\text{mass of Ca}}{\text{molar mass of Ca}} = \frac{10.00 \text{ g}}{40.08 \text{ g/mol}} = 0.249 \text{ mol}$$
- Step 2: The mole ratio of V$_2$O$_5$ and Ca from the balanced equation is 1:5, which means that 1 mol of V$_2$O$_5$ reacts with 5 mol of Ca. However, the mole ratio of V$_2$O$_5$ and Ca from the given masses is 0.0550:0.249, which is equivalent to 1:4.53. This means that there is not enough Ca to react with all of the V$_2$O$_5$. Therefore, Ca is the limiting reagent and V$_2$O$_5$ is the excess reagent.
- Step 3: The mole ratio of Ca and CaO from the balanced equation is 5:5, which means that 5 mol of Ca produce 5 mol of CaO. The molar mass of CaO is 56.08 g/mol. Therefore, the theoretical yield of CaO is:
$$\text{theoretical yield of CaO} = \text{moles of Ca} \times \frac{\text{moles of CaO}}{\text{moles of Ca}} \times \text{molar mass of CaO}$$
$$\text{theoretical yield of CaO} = 0.249 \text{ mol} \times \frac{5 \text{ mol}}{5 \text{ mol}} \times 56.08 \text{ g/mol}$$
$$\text{theoretical yield of CaO} = 13.96 \text{ g}$$
- Step 4: The percent yield of CaO is:
$$\text{percent yield of CaO} = \frac{\text{actual yield of CaO}}{\text{theoretical yield of CaO}} \times 100\%$$
$$\text{percent yield of CaO} = \frac{10.53 \text{ g}}{13.96 \text{ g}} \times 100\%$$
$$\text{percent yield of CaO} = 75.4\%$$
- Step 5: The amount of the limiting reagent (Ca) that remains after the reaction is zero, since it is completely consumed. The amount of the excess reagent (V$_2$O$_5$) that remains after the reaction can be calculated by subtracting the amount of V$_2$O$_5$ that reacted with Ca from the initial amount of V$_2$O$_5$. The mole ratio of V$_2$O$_5$ and Ca from the balanced equation is 1:5, which means that 1 mol of V$_2$O$_5$ reacts with 5 mol of Ca. Therefore, the amount of V$_2$O$_5$ that remains is:
$$\text{amount of V}_2\text{O}_5 \text{ that remains} = \text{initial amount of V}_2\text{O}_5 - \text{amount of V}_2\text{O}_5 \text{ that reacted}$$
$$\text{amount of V}_2\text{O}_5 \text{ that remains} = 0.0550 \text{ mol} - \frac{\text{moles of Ca}}{\frac{\text{moles of Ca}}{\text{moles of V}_2\text{O}_5}}}$$
$$\text{amount of V}_2\text{O}_5 \text{ that remains} = 0.0550 \text{ mol} - \frac{0.249 \text{ mol}}{5}$$
$$\text{amount of V}_2\text{O}_5 \text{ that remains} = 0.0050 \text{ mol}$$
To convert this to grams, we multiply by the molar mass of V$_2$O$_5$:
$$\text{amount of V}_2\text{O}_5 \text{ that remains} = 0.0050 \text{ mol} \times 181.88 \text{ g/mol}$$
$$\text{amount of V}_2\text{O}_5 \text{ that remains} = 0.91 \text{ g}$$
Here are the answers to your questions:
- A) The limiting reagent is Ca.
- B) The theoretical yield of CaO is 13.96 g.
- C) The amount of the excess reagent (V$_2$O$_5$) that remains is 0.91 g.
- D) The percent yield of CaO is 75.4%.
- E) The amount of the limiting reagent (Ca) that remains is zero.