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

Assuming a 100% yield, 11.1 g of acrylonitrile can be produced from the given masses of reactants.

Step-by-step solution

Calculate the moles of reactants

First, we need to convert the mass of each reactant to moles. We can do this by dividing the mass of each reactant by its molar mass. The molar mass of C3H6 is: \(12.01 \times 3 + 1.01 \times 6 = 42.09 \, g/mol\) The molar mass of O2 is: \(16.00 \times 2 = 32.00 \, g/mol\) The molar mass of NH3 is: \(14.01 + 1.01 \times 3 = 17.04 \, g/mol\) Moles of C3H6 = \(\frac{15.0 \, g}{42.09 \, g/mol} = 0.356 \, mol\) Moles of O2 = \(\frac{10.0 \, g}{32.00 \, g/mol} = 0.313 \, mol\) Moles of NH3 = \(\frac{5.00 \, g}{17.04 \, g/mol} = 0.294 \, mol\)

Determine the limiting reactant

The limiting reactant is the reactant that will be completely consumed first, based on stoichiometry. We will compare the mole ratios of the reactants to determine the limiting reactant. For C3H6: \(\frac{0.356\, mol}{2} = 0.178\) For O2: \(\frac{0.313\, mol}{3} = 0.104\) For NH3: \(\frac{0.294\, mol}{2} = 0.147\) Since the value for O2 is the smallest, it is the limiting reactant.

Calculate the moles of acrylonitrile produced

Using the stoichiometry provided in the balanced chemical equation, we can calculate the moles of acrylonitrile produced based on the moles of the limiting reactant (O2). Moles of acrylonitrile (C3H3N) = moles of O2 × \(\frac{2 \, mol \, C3H3N}{3 \, mol \, O2}\) = \(0.313 \, mol \times \frac{2}{3} = 0.209 \, mol\)

Calculate the mass of acrylonitrile produced

Now that we have the moles of acrylonitrile produced, we can calculate its mass by multiplying the moles with its molar mass. The molar mass of C3H3N is: \(12.01 \times 3 + 1.01 \times 3 + 14.01 = 53.06 \, g/mol\) Mass of acrylonitrile = \(0.209 \, mol \times 53.06 \, g/mol = 11.1 \, g\) Therefore, assuming a 100% yield, 11.1 g of acrylonitrile can be produced from the given masses of reactants.

Key concepts

Limiting Reactant

Understanding the concept of the limiting reactant is critical for predicting how much product can be formed in a chemical reaction. The limiting reactant is the substance that is entirely used up first during a chemical reaction, controlling the extent to which the reaction can occur.

In our exercise, we've measured the amounts of propane (C₃H₆), oxygen (O₂), and ammonia (NH₃) and compared the amount of moles we have to what the balanced chemical equation requires. It's like a chef who has to make a certain number of sandwiches but has a limited amount of bread — no matter how much of the other ingredients are available, the number of sandwiches that can be made is limited by the amount of bread. Similarly, the amount of product we can make in a chemical reaction is limited by the limiting reactant, in this case, oxygen (O₂).

To improve comprehension, it's helpful to imagine that you're building something with Lego blocks. If you run out of red blocks required for your model, it doesn't matter how many blue or yellow blocks you have left; you can only make as many complete models as the number of red blocks allows.

Chemical Reaction Yield

The yield of a chemical reaction is an important aspect that denotes the efficiency of the reaction. Theoretical yield is the number of products that can be made based on stoichiometry, assuming everything goes perfectly and no product is lost. However, real-life chemistry isn't perfect. The actual yield is usually less than the theoretical yield due to side reactions, incomplete reactions, or loss of product during the process.

The actual yield is stated as a percentage of the theoretical yield, known as the percent yield. Even if a reaction is assumed to have a 100% yield for simplicity, as in our exercise, being aware that it's an idealization is crucial. For more practical understanding, compare the yield to baking cookies with a fixed recipe. The theoretical yield represents the maximum number of cookies you could get from the dough, but the actual yield might be less if some of the dough sticks to the bowl or gets burnt in the oven.

Molar Mass Calculation

Molar mass calculation is a fundamental skill in chemistry that allows us to convert between mass and moles of a substance. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To calculate the molar mass of a compound, add up the molar masses of each element in the compound, multiplied by the number of atoms of that element in a molecule.

In the exercise, calculating the molar mass allowed us to understand how much product could form from a given mass of reactants. It's like understanding the weight of a 'bunch' of different items. If you know the weight of one apple (like the molar mass of one element), then you can easily figure out how much a 'bunch' of apples (or a mole of molecules) weighs by multiplying by the correct number of apples in the 'bunch' (or atoms in a molecule).