2.2. Billions of kilograms of urea, are produced annually for use as a fertilizer. The reaction used is given below.
NH3(g) + CO2(g) → CO(NH2)2(s) + H2O(l)
The typical starting reaction mixture has a 3:1 mole ratio of NH3 to CO2. If 47.7 g urea forms per mole of that reacts, what is the
2.2.1. Theoretical yield; [5]
2.2.2. Actual yield; [2]
2.2.3. Percent yield? [2]
To solve for the theoretical yield, actual yield, and percent yield, we first need to calculate the number of moles of NH3 and CO2 in the reaction mixture.
Let's start by finding the moles of NH3. We know that the starting mixture has a 3:1 mole ratio of NH3 to CO2. Therefore, if x is the number of moles of CO2, the number of moles of NH3 would be 3x.
Next, we can convert the given mass of urea (47.7 g) to moles using the molar mass of urea. The molar mass of urea (CO(NH2)2) is 60.06 g/mol. So, the number of moles of urea formed would be 47.7 g / 60.06 g/mol.
Now that we have the number of moles of NH3, CO2, and urea, we can answer the questions.
2.2.1. Theoretical Yield:
The theoretical yield is the maximum amount of urea that can be formed according to the balanced chemical equation. The balanced equation tells us that 1 mole of NH3 reacts to form 1 mole of CO(NH2)2. Therefore, the theoretical yield would be equal to the number of moles of urea formed, which we calculated earlier.
2.2.2. Actual Yield:
The actual yield is the amount of urea actually obtained from the reaction. This value is typically determined through experimentation or given in the question.
2.2.3. Percent Yield:
The percent yield is a measure of the efficiency of the reaction and is calculated by dividing the actual yield by the theoretical yield, then multiplying by 100. The formula for percent yield is:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Finally, by substituting the values of the actual and theoretical yields, we can calculate the percent yield.
Please provide the value for the actual yield in order to calculate the percent yield.