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Chapter 3: Problem 128

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

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

Step by step solution

01

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
02

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.
03

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.

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Most popular questions from this chapter

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 -g sample of impure iron ore is heated with excess carbon, producing 453 g of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is 100\(\%\) efficient.

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$ \begin{array}{l}{\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\\ {\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)}\end{array} $$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Methane \(\left(\mathrm{CH}_{4}\right)\) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. a. Write the balanced chemical equation for the reaction of methane and sulfur. b. Calculate the theoretical yield of carbon disulfide when \(120 .\) g of methane is reacted with an equal mass of sulfur.

A \(0.4230-\) g sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Boron consists of two isotopes, 10 \(\mathrm{B}\) and 11 \(\mathrm{B}\) . Chlorine also has two isotopes, 35 \(\mathrm{Cl}\) and 37 \(\mathrm{Cl}\) . Consider the mass spectrum of \(\mathrm{BCl}_{3}\) . How many peaks would be present, and what approximate mass would each peak correspond to in the BCl_ mass spectrum?
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