Question

determination	1	2
mass of Zn strip, g	25.469	23.887
mass of empty evaporating dish, g	23.274	20.740
mass of dried unreated Zn, g	24.678	23.372
mass of evaportating dish plus Cu, g (first heating)	24.741	24.499
second heating	24.066	21.240
third heating	23.980	21.240

mass of copper chloride in 25.0 ml of solution, g = 2.018g

PLEASE SHOW ALL WORK AND BE AS DESCRIPTIVE AS POSSIBLE PLEASE AND THANK YOU!

1. (A) Calculate the mass of Cu produced in both determinations.

determination 1:_______ determination 2:________

(B) Calculate the mass of chlorine (Cl) in your original Cu(x)Cl(y) sample.

determination 1:_______ determination 2:________

which of the reactants was the limiting factor? Which reactant was in excess?

(C) Calculate the number of moles of Cu present in your Cu(x)Cl(y) sample and calculate the number of moles of Cl present in the Cu(x)Cl(y) sample.

mole of Cu determination 1:_______ determination 2:________

moles of Cl determination 1:_______ determination 2:________

(D) Determine the molar ratio of Cu to Cl in Cu(x)Cl(y) then use the ration to write the empirical formula for copper chloride.

determination 1:_______ determination 2:________

(E) Determine the pecent of Cu and the percent of Cl in copper(II) chloride using the formula CuCl2

(F) Using the mass of Zn reacted in each determination calculate the theoretical yield of Cu for each using the below equation

theoretical yield of Cu = (mass of Zn reacted) x 1molZn/65.38g Zn x 1mol Cu/1mol Zn x 63.55g Cu/1mol Cu

Finally determine the percent yield from each of the determinations for Cu using the following equation

percent yield of Cu= actual mass of Cu produced/ theoretical yield of Cu X 100

If the yeild is not 100% explain why it is not.

PLEASE SHOW ALL WORK AND BE AS DESCRIPTIVE AS POSSIBLE PLEASE AND THANK YOU!

Answer 1

As mentioned in comment you want only answers for #2 .

A ) Mass of copper produced = mass of evaportating dish plus Cu - Mass of evaporating dish

Mass of copper = 21.24 - 20.74 = 0.5 grams

B) Mass of chloride in copper chloride = Mass of copper chloride - Mass of copper = 2.018 - 0.5 = 1.518 grams

C) Moles of Copper present = Mass of copper / atomic weight of copper = 0.5 / 63.5 = 0.00787

Moles of chloride = Mass / Atomic weight = 1.518 / 35.5 = 0.0427

The formulat should be CuCl5 (not possible however , there is some mistake in the mass of copper chloride given)

E) the % in CuCl2 will be

Total mass = 63.5 + 35.5 + 35.5 = 134.5

Mass of Cu = 63.5 so % of Cu = 63.5 / 134.5 = 47.2 %

So % of chlorine = 52.8%

F) theoretical yield of Cu = (mass of Zn reacted) x 1molZn/65.38g Zn x 1mol Cu/1mol Zn x 63.55g Cu/1mol Cu

Mass of Zn reacted = Mass of Zn initially taken - Mass of Zn left =23.887 - 23.372 = 0.515 grams

Theoretical yield = (0.515 X 63.55 / 65.38 ) = 0.5 grams

percent yield of Cu= actual mass of Cu produced/ theoretical yield of Cu X 100

% yield of Cu = 0.5 X 100 / 0.5 = 100%