Question

Acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. $$ 2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(15.0 \mathrm{g} \mathrm{C}_{3} \mathrm{H}_{6}, 10.0 \mathrm{g} \mathrm{O}_{2},\) and 5.00 \(\mathrm{g} \mathrm{NH}_{3}\) are reacted, what mass of acrylonitrile can be produced, assuming 100\(\%\) yield?

Short answer

The mass of acrylonitrile that can be produced, assuming 100% yield, is \(11.0 \mathrm{g}\).

Step-by-step solution

1. Calculate the number of moles for each reactant

First, we'll find the molar masses of the given reactants: \(\mathrm{C}_{3} \mathrm{H}_{6}, \mathrm{O}_{2},\) and \(\mathrm{NH}_{3}\). Next, we'll convert the given mass of each reactant to moles by dividing the mass by its molar mass. Molar masses: \(C_3H_6 = (3 * 12.01) + (6 * 1.01) = 42.09\) g/mol \(O_2 = (2 * 16.00) = 32.00\) g/mol \(NH_3 = (14.01) + (3 * 1.01) = 17.03\) g/mol Converting mass to moles: Moles of \(C_3H_6 = \frac{15.0\text{ g}}{42.09 \text{ g/mol}} = 0.356\) moles Moles of \(O_2 = \frac{10.0\text{ g}}{32.00 \text{ g/mol}} = 0.313\) moles Moles of \(NH_3 = \frac{5.00\text{ g}}{17.03 \text{ g/mol}} = 0.294\) moles

2. Determine the limiting reactant

Using the balanced chemical equation, we can see that the mole ratio of reactants is \(2:2:3\) for \(C_3H_6:NH_3:O_2\). We'll now divide the number of moles of each reactant by their respective coefficients in the balanced equation to find the limiting reactant. \(\frac{\text{moles of } C_3H_6}{2} = \frac{0.356}{2} = 0.178\) \(\frac{\text{moles of } NH_3}{2} = \frac{0.294}{2} = 0.147\) \(\frac{\text{moles of } O_2}{3} = \frac{0.313}{3} = 0.104\) Since 0.104 is the smallest value, \(O_2\) is the limiting reactant.

3. Calculate the mass of acrylonitrile produced

Now, we'll use the mole ratio between the limiting reactant and the product (acrylonitrile) to find the moles of acrylonitrile produced. The mole ratio is \(3:2\) for \(O_2:C_3H_3N\). Moles of acrylonitrile produced = \(\frac{2}{3} \times 0.313\) moles of \(O_2 = 0.208\) moles Finally, we'll convert the moles of acrylonitrile to mass. The molar mass of \(C_3H_3N\) is: \((3 * 12.01) + (3 * 1.01) + (14.01) = 53.06\) g/mol. Mass of acrylonitrile produced = \(0.208 \text{ moles} \times 53.06 \text{ g/mol} = 11.0 \mathrm{g}\) (rounded to one decimal place) The mass of acrylonitrile that can be produced, assuming 100% yield, is 11.0 g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is entirely consumed first, stopping the reaction from continuing and limiting the amount of product that can be formed. To identify the limiting reactant, we need to compare the mole ratios of the reactants used against their coefficients in the balanced chemical equation.
This is essential because reactants are rarely provided in the exact proportions needed for the reaction. Therefore, one reactant runs out first, determining the maximum capacity of product formation.

• Start by calculating moles for each reactant given their initial mass and molar mass.
• Then, use the coefficients from the balanced equation to calculate how many moles of product each reactant can produce.
• The reactant that produces the least amount of product is your limiting reactant.

In our example, we found that oxygen (\(O_2\)) was the limiting reactant because it yielded the smallest value when calculations were completed for all reactants. Understanding limiting reactants is crucial for realistic and efficient industrial chemical processes.

Mole Conversions

Mole conversions serve as a bridge between the mass of a substance and the number of molecules or atoms it represents. In chemical reactions, working with moles is essential because they tie the quantities of substances to equivalent amounts, allowing for straightforward stoichiometry.
In the exercise, mole conversions were carried out using the molar masses of each substance. Here's how it's done:

• Calculate the molar mass by summing the atomic masses of all atoms in a molecule.
• Use the formula: moles = mass / molar mass.

For example, the molar mass of \(C_3H_6\) was calculated as \(42.09\) g/mol. With a mass provided, we converted it into moles to understand how much of \(C_3H_6\) was available for the reaction. These conversions are the backbone of stoichiometric calculations, making it possible to determine quantities of other substances in the reaction and ultimately the products formed.

Chemical Reactions

Chemical reactions involve the transformation of reactants through rearrangement of atoms resulting in the formation of products. This is governed by a balanced chemical equation which shows the precise quantity relationships.
In the example provided, acrylonitrile is formed through a specific reaction involving propane (\(C_3H_6\)), ammonia (\(NH_3\)), and oxygen (\(O_2\)) to form acrylonitrile (\(C_3H_3N\)) and water (\(H_2O\)).

• Reactions proceed according to the mole ratio as dictated by the coefficients in the balanced equation.
• Balanced equations must satisfy conservation of mass - meaning the number of atoms for each element must be the same on both sides of the equation.

In industrial and laboratory settings, the balanced chemical equation not only dictates how much product can be obtained but also how safely and efficiently a process can be carried out. Understanding chemical reactions on this level is key to manipulating them for desired outcomes, such as producing acrylonitrile for synthetic carpets and fabrics.