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

a.

moles of propane present before the reaction = 5.0 / 44.1 = 0.113 mole

b.

mass of Br2 = density * volume = 3.10 * 10 = 31.0 g

mole of Br2 = 31.0/159.8 = 0.194 mol

c.

each propane molecule need two molecules of bromine.

mole of Br2 required for 0.113 mol propae = 0.113 * 2 = 0.226 mol

as mole of Br2 is present in less than required amount therefore bromine is limiting reagent.

d.

mole of excess propane = 0.113 - 0.194/2 = 0.016 mol

mass of excess reactant = 0.016*44.1 = 0.706 g

e.

theoretical yield = 0.194/2 * 201.9 = 19.58 g

f.

mass of 2,2 dibrompropane = density * volume = 1.78 * 15.58 = 27.7 g

mole of 2,2 dibrompropane = 15.58/201.9 = 0.0772 mol

mmol of HCL = 3.0 * 10^-6 * 12 *1000 = 0.036 mmol

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