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?

To convert grams to moles, we will use the molar masses of the given substances. Molar mass of C3H6: \(3 \times 12.01 \, g/mol (C) + 6 \times 1.01 \, g/mol (H) = 42.08 \,g/mol\) Molar mass of O2: \(2 \times 16.00 \, g/mol = 32.00 \, g/mol\) Molar mass of NH3: \(1.01 \times 3 \, g/mol (H) + 14.01 \, g/mol(N) = 17.03 \,g/mol\) Now, we will convert the given grams of reactants to moles using their respective molar masses: Moles of C3H6: \(\frac{15.0 \,g}{42.08 \, g/mol} = 0.356 \, mol\) Moles of O2: \(\frac{10.0 \,g}{32.00 \, g/mol} = 0.313 \, mol\) Moles of NH3: \(\frac{5.00 \,g}{17.03 \, g/mol} = 0.294 \, mol\)

Using the balanced chemical equation, we can determine the mole ratios of the reactants. The balanced equation is: \(2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) Using the given moles of reactants, we can find out the possible moles of Acrylonitrile that can be produced. For C3H6: \(\frac{0.356 \, mol \, C_{3}H_{6}}{2} = 0.178 \, mol \, C_{3}H_{3}N\) For O2: \(\frac{0.313 \, mol \, O_{2}}{3} = 0.104 \, mol \, C_{3}H_{3}N\) For NH3: \(\frac{0.294 \, mol \, NH_{3}}{2} = 0.147 \, mol \, C_{3}H_{3}N\) Since the reaction can only proceed until the limiting reactant is used up, the smallest possible moles of Acrylonitrile will determine the limiting reactant. In this case, O2 is the limiting reactant.

Using the limiting reactant O2, we can now calculate the mass of Acrylonitrile that can be produced. First, we need to find the molar mass of C3H3N. Molar mass of C3H3N: \(3 \times 12.01 \, g/mol (C) + 3 \times 1.01 \, g/mol (H) + 1 \times 14.01 \, g/mol (N) = 53.06 \, g/mol\) Now, we will convert the possible moles of Acrylonitrile (C3H3N) produced from O2 into grams: Mass of C3H3N: \(0.104 \, mol \, C_{3}H_{3}N \times 53.06 \, g/mol = 5.52 \, g\) Assuming 100% yield, 5.52 grams of Acrylonitrile can be produced from the given amounts of reactants.