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Chapter 3: Problem 122

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
Assuming 100% yield, the mass of acrylonitrile produced is approximately \(11.09\ \mathrm{g}\).

Step by step solution

01

Calculate the moles of each reactant

We should determine the number of moles for each reactant by dividing their mass by their corresponding molar mass. The molar mass of C3H6 is \( 3(12.01) + 6(1.01) \approx 42.08 \ g/mol \), for O2 is \( 2(16.00) = 32.00 \ g/mol \), and for NH3 is \( 14.01 + 3(1.01) \approx 17.03 \ g/mol \). Now we can find the moles of each reactant. moles of C3H6: \(\frac{15.0\mathrm{~g}}{42.08\mathrm{~g/mol}} \approx 0.356\ \mathrm{mol}\) moles of O2: \(\frac{10.0\mathrm{~g}}{32.00\mathrm{~g/mol}} \approx 0.313\ \mathrm{mol}\) moles of NH3: \(\frac{5.00\mathrm{~g}}{17.03\mathrm{~g/mol}} \approx 0.294\ \mathrm{mol}\)
02

Determine the limiting reactant

The stoichiometry of the reaction is: \(2 \ C_3H_6 + 2 \ NH_3 + 3 \ O_2 \longrightarrow 2 \ C_3H_3N + 6 \ H_2O\) We must divide the moles of each reactant by their stoichiometric coefficient and pick the one with the smallest ratio. \[\frac{0.356}{2} \approx 0.178\] \[\frac{0.294}{2} \approx 0.147\] \[\frac{0.313}{3} \approx 0.104\] The smallest value is \(0.104\), so the limiting reactant is O2.
03

Calculate the mass of acrylonitrile produced

Now we can use the stoichiometry of the reaction to determine the moles of C3H3N produced: \(2 \ C_3H_6 + 2 \ NH_3 + 3 \ O_2 \longrightarrow 2 \ C_3H_3N + 6 \ H_2O\) For every 3 moles of O2 consumed, 2 moles of C3H3N are produced. moles of C3H3N: \(\frac{2}{3} × 0.313 \approx 0.209\ \mathrm{mol}\) Now we can convert moles of C3H3N back to grams by multiplying with the molar mass of C3H3N (\(3(12.01) + 3(1.01) + 14.01 \approx 53.07 g/mol\)): mass of C3H3N: \(0.209\ \mathrm{mol} × 53.07\ \mathrm{g/mol}\approx 11.09\ \mathrm{g}\) Assuming 100% yield, the mass of acrylonitrile produced is approximately\( \boxed{11.09\ \mathrm{g}}\).

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Most popular questions from this chapter

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating \(0.6498 \mathrm{~g}\) of one of the compounds leaves a residue of \(0.6018 \mathrm{~g}\). Heating \(0.4172 \mathrm{~g}\) of the other compound results in a mass loss of \(0.016 \mathrm{~g}\). Determine the empirical formula of each compound.

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