consider the reaction of 5.0g of propane with 10.0ml ofmolecular bromine to form

Question

consider the reaction of 5.0g of propane with 10.0ml ofmolecular bromine to form 2,2 dibromopropane and hydrogenbromide. a. look up the molecular weight of propane (44.1) andcalculate the moles of propane present before the reaction b. look up the density of and molecular weight ofbromine and calculate the moles of molecula bromine present beforethe reaction c. Determine whether propane or bromine is the limitingreactant d. calculate the mass (g) of the excess reactant after thereaction is complete e. calculate the theoretical yeild (g) of 2,2dibromopropane f. suppose that 15.58g of 2,2 dibromopropane were actuallyformed in the reaction. Calculate the percent yield of2,2-dibromopropane. g. look up the density of 2,2 dibrompropane and calculate thevolume of 15.58g of dibromopropane. 2. Calculate the number of mmol of HCL in 3.0ul of 12 MHCL consider the reaction of 5.0g of propane with 10.0ml ofmolecular bromine to form 2,2 dibromopropane and hydrogenbromide. a. look up the molecular weight of propane (44.1) andcalculate the moles of propane present before the reaction b. look up the density of and molecular weight ofbromine and calculate the moles of molecula bromine present beforethe reaction c. Determine whether propane or bromine is the limitingreactant d. calculate the mass (g) of the excess reactant after thereaction is complete e. calculate the theoretical yeild (g) of 2,2dibromopropane f. suppose that 15.58g of 2,2 dibromopropane were actuallyformed in the reaction. Calculate the percent yield of2,2-dibromopropane. g. look up the density of 2,2 dibrompropane and calculate thevolume of 15.58g of dibromopropane. 2. Calculate the number of mmol of HCL in 3.0ul of 12 MHCL

Explanation / Answer

I know this reply is too late to help you much (feel free to give alow rating), but here's how to do it, because you'll probably needto be able to do these sorts of calculations again in thefuture: First, write a balanced reaction scheme: Propane + bromine --> 2,2-dibromopropane + hydrogen bromide C3H8   +   2Br2  -->           C3H6Br2       +          2HBr (A) MW(C3H8) = 44.0956 g/mol      n(C3H8) = mass /molar mass                   = 5.0 g / 44.0956 g/mol (alwaysput units in your calculations! it will let you know if you'redoing them right!)                   = 0.1134 mol (B) Density(Br2) = 3.03 g/mL     MW(Br2) = 159.808 g/mol          n(Br2) = mass / molar mass = (vol * density) / molarmass                    = 10.0 mL * 3.03g/mL / 159.808 g/mol                    = 0.1896 mol (C) We have less moles of propane than bromine, so propane is thelimiting reagent. (D) The excess of bromine is:     n(Br2) -n(C3H8) = 0.1896 - 0.1134 = 0.0762 mol    mass(excess Br2) = 0.0762mol *159.808 g/mol = 12.2 g (E) One mole of propane produces one mole of 2,2-dibromopropane(DBP), so:     n(DBP) = n(propane) = 0.1134 mol     MW(DBP) = 201.8877 g/mol     mass(DBP) = 0.1134 mol *201.8877 g/mol = 22.89 g (F)     % Yield = Actual / theoretical yield * 100%                 = 15.58 g / 22.89 g * 100%                 = 68% (G)     Density(DBP) = 1.780g/mL          vol(DBP) = 15.58 g / 1.780 g/mL =8.75 mL 2. First convert uL to L:     3 uL = 3/1000 mL = 3/1000000 L =3x10-6 L Now use the relationship: concentration = moles / volume tocalculate the moles of HCl:     conc = mol / vol mol = conc * vol            = 12mol/L * 3x10-6L =3.6x10-5 mol Finally, convert mol to mmol:     3.6x10-5 mol * 1000mmol/mol = 0.036 mmol

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