Question

Each step in the process below has a 70.0% yield.

CH4+4Cl2-->CCl4+4HCl

CCl4+2HF-->CCl2F2+2HCl

The CCl4 formed in the first step is used as a reactant in the second step. If 7.00 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

Answer 1

The CCl4 formed in the first step is used as a reactant in the second step. If 7.00 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

If 7 mole CH4 reacts in the first stage, you will get 0.7 x 4 x 7 moles HCl(=19.6moles) in the first stage and 0.7x 7 moles of CCl4 (=4.9 mole)

In the second stage each mole CCl4 makes 2 mole HCl at 100% yield so with a 70% yield and starting with 4.9 mole CCl4 you will get 0.7 x 2 x 4.9 mole HCl(= 6.86 moles). Add this to the amount in the first stage and you have the total moles HCl made

total moles HCl = total moles HCl in first step + total moles HCl in second step

                        = 19.6 + 6.86

                        = 26.46 moles

total mol HCl produced is 26.46 mol

OR

moles of given chemical x molar ratio from balance equation:

this ratio is in the following form: unknown coeff./know coeff.

so 7 mol. CH4 x 4 HCl/1 CH4 = 28 mol HCl in step 1 x 0.7 = 19.6 mol HCl

also in step one figure out how much CCl4 is made: 7 mol CH4 x 1 mol CCl4/1mol CH4 = 7 Mol CCl4 produced x 0.7 = 4.9 mol CCl4

With 4.9 mole of CCl4 produced to use in step 2, how much more HCl is produced? 4.9 mol CCl4 x 2mol HCl/1 mole CCl4 = 9.8 mole HCl x 0.7 =6.86 mol HCl Produced

Add the 19.6 mol HCl in step 1 to the 6.86 mol HCl in step 2 and total mol HCl produced is 26.46 mol HCl totally produced.