Question

1. The percent yield for the reaction PCl3 + Cl2 → PCl5 is 62.6 percent. What mass of PCl5 would be expected from the reaction of 32.1 grams of PCl3 with excess chlorine?

2. Balance the chemical equation for the important industrial process: C2H4 + O2 + HCl → C2H4Cl2 + H2O. Given 675.0 kg of C2H4, calculate how many kg of C2H4Cl2 can be produced.

3. Calculate the number of moles in 2.75 μg of Te.

Answer 1

Q1.

get moles of PCl3

mol = mass/MW = 32.1/137,3328 = 0.23373 mol of PCl3 present

now, ratio is

1 mol of PCl3 = 1 mol of PCl5

0.23373 mol of PCl3 = 0.23373 mol of PCl5

mass = mol*MW = 0.23373*208.2388 = 48.6716 g of PCl5 expected

now,

at 62.6% yield --> 0.626 *  48.6716 = 30.468 g only of PCl5 will be produced

Q2

C2H4 + O2 + HCl → C2H4Cl2 + H2O

balance

2 C2H4 + O2 + 4 HCl = 2 C2H4Cl2 + 2 H2O

now

mol of C2H4 = mass/MW = 675/28.0532 = 24.0614 kmol

ratio is 2 mol of C2H4 = 2 mol

C2H4 24.0614 kmol = 24.0614 kmol of C2H4Cl2

mass of C2H4Cl2 = mol*MW C2H4Cl2 = (24.0614 )(98.9592 ) = 2381.09 kg

Q3

maol inf

m = 2.75 microg of Te

m = 2.75*10^-6 g of Te

MW of Te = 127.6 g/mol

mol = mass/MW = (2.75*10^-6)/(127.6 ) = 2.1551*10^-8 moles of Te