Grade 10

Advanced

10 min

Lesson 12 of 12

Advanced Stoichiometry Applications

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Synthesizing all concepts to solve complex, multi-variable problems found in industrial chemistry.

Imagine you are the lead engineer at a multi-billion dollar pharmaceutical plant; if your calculations are off by just 1%, the company loses millions and the medicine supply fails. How do you ensure every atom is accounted for in a massive industrial reaction?

Learning Objectives

1.Solve integrated problems combining mass, gas volume, and percent yield simultaneously
2.Evaluate the cost-effectiveness of industrial reactions based on stoichiometric efficiency
3.Create a step-by-step procedure for calculating reagents needed for a specific industrial output

The Industrial Mole Bridge

In laboratory settings, we often deal with single variables. However, in Industrial Chemistry, we must synthesize multiple concepts at once. The core of every problem remains the Mole Bridge. Whether you start with a gas volume ( $V$ ), a solution concentration ( $M$ ), or a solid mass ( $m$ ), you must first convert to moles ( $n$ ) before using the coefficients of a balanced equation to 'cross over' to your product. In industry, we use the Ideal Gas Law ( $PV = nRT$ ) and Molar Volume ( $22.4 \text{ L/mol}$ at STP) to bridge the gap between gaseous reactants and solid products.

How many grams of Lithium Hydroxide ( $LiOH$ ) are needed to remove $500 \text{ L}$ of $CO_2$ gas at STP from a spacecraft's atmosphere? 1. Write the balanced equation: $2LiOH(s) + CO_2(g) \rightarrow Li_2CO_3(s) + H_2O(l)$ 2. Convert $CO_2$ volume to moles: $n = \frac{500 \text{ L}}{22.4 \text{ L/mol}} = 22.32 \text{ mol}$ 3. Use the molar ratio ( $2:1$ ): $22.32 \text{ mol } CO_2 \times \frac{2 \text{ mol } LiOH}{1 \text{ mol } CO_2} = 44.64 \text{ mol } LiOH$ 4. Convert moles to mass: $44.64 \text{ mol} \times 23.95 \text{ g/mol} = 1069.1 \text{ g}$

Quick Check

If you are given the volume of a gas at STP, what constant do you use to find the number of moles?

Think about the volume one mole of any gas occupies at standard temperature and pressure.

Answer

The molar volume of a gas at STP, which is $22.4 \text{ L/mol}$ .

The Efficiency Tax: Yield and Purity

Real-world chemistry is rarely perfect. Two factors 'tax' our theoretical results: Percent Purity and Percent Yield. If an industrial ore is only

80\%

pure, only

0.80

of every gram is actual reactant. Similarly, Percent Yield accounts for product lost during filtration or incomplete reactions. The formula is:

\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100

In complex problems, you must apply purity at the beginning (reactants) and yield at the end (products).

An industrial plant produces Ammonia ( $NH_3$ ) from Nitrogen gas. If they start with $100 \text{ kg}$ of $N_2$ and the process has an $85\%$ yield, what is the actual mass of $NH_3$ produced? 1. Equation: $N_2 + 3H_2 \rightarrow 2NH_3$ 2. Moles of $N_2$ : $\frac{100,000 \text{ g}}{28.02 \text{ g/mol}} = 3568.87 \text{ mol}$ 3. Theoretical moles of $NH_3$ : $3568.87 \times 2 = 7137.74 \text{ mol}$ 4. Theoretical mass: $7137.74 \text{ mol} \times 17.03 \text{ g/mol} = 121,555.7 \text{ g} \approx 121.56 \text{ kg}$ 5. Apply yield: $121.56 \text{ kg} \times 0.85 = 103.33 \text{ kg}$

Quick Check

If a reaction has a theoretical yield of $50 \text{ g}$ but you only collect $40 \text{ g}$ , what is the percent yield?

Divide the actual amount by the predicted amount.

Answer

$80\%$ . Calculation: $(40 / 50) \times 100 = 80\%$ .

Economic Stoichiometry

In industry, the Limiting Reagent is chosen strategically. We usually make the most expensive chemical the limiting reagent to ensure it is completely consumed, while using a cheaper chemical in excess. To evaluate cost-effectiveness, engineers calculate the Unit Cost of Product:

\text{Cost per gram} = \frac{\text{Total Cost of Reactants}}{\text{Actual Mass of Product}}

. Minimizing waste and maximizing yield directly dictates the profit margins of chemical corporations.

You need to produce $1000 \text{ kg}$ of Titanium ( $Ti$ ) via the Kroll Process: $TiCl_4 + 2Mg \rightarrow Ti + 2MgCl_2$ . The process has a $90\%$ yield. $Mg$ costs $\$ 5.00/kg $. Calculate the cost of$ Mg $needed. 1. Target actual yield:$ 1000 \text{ kg} $. 2. Calculate required theoretical yield:$ \frac{1000 \text{ kg}}{0.90} = 1111.11 \text{ kg } Ti $. 3. Moles of$ Ti $needed:$ \frac{1,111,110 \text{ g}}{47.87 \text{ g/mol}} = 23,211 \text{ mol} $. 4. Moles of$ Mg $required ($ 2:1 $ratio):$ 23,211 \times 2 = 46,422 \text{ mol} $. 5. Mass of$ Mg $:$ 46,422 \text{ mol} \times 24.31 \text{ g/mol} = 1,128,518 \text{ g} \approx 1128.5 \text{ kg} $. 6. Total Cost:$ 1128.5 \text{ kg} \times \ $5.00 = \$ 5,642.50$.

Key Takeaways

1The Mole Bridge is the central transition point for all stoichiometry calculations, regardless of the state of matter.
2Percent Yield and Percent Purity act as multipliers (e.g., $0.85$ for $85\%$ ) that reduce the final output of a reaction.
3Industrial efficiency is measured by the cost-effectiveness of reagents relative to the actual yield of the product.
4Always convert target actual yields back to theoretical yields before using molar ratios.

Quick Check

Multiple Choice

In an industrial process, if the 'Actual Yield' is $200 \text{ kg}$ and the 'Percent Yield' is $80\%$ , what was the 'Theoretical Yield'?

Multiple Choice

Which step must be performed FIRST when calculating the mass of product from a given volume of gas at STP?

True/False

Using a reagent in excess is a common industrial strategy to ensure the more expensive reagent is fully consumed.

What's Next?

Review Tomorrow

In 24 hours, try to recall the three-step sequence for solving a yield-adjusted stoichiometry problem: Purity → Mole Bridge → Yield.

Practice Activity

Find a recipe for cookies and calculate the 'Percent Yield' by comparing the number of cookies the recipe says it makes vs. how many you actually produce.

Browse

Browse

Advanced Stoichiometry Applications

Learning Objectives

The Industrial Mole Bridge

The Efficiency Tax: Yield and Purity

Economic Stoichiometry

Key Takeaways

Quick Check

What's Next?

Advanced Stoichiometry Applications

Learning Objectives

The Industrial Mole Bridge

The Efficiency Tax: Yield and Purity

Economic Stoichiometry

Key Takeaways

Quick Check

What's Next?