What is the theoretical yield of Copper Oxide with Sulfuric Acid?

Question

Equation: 
CuO + H2SO4 --> CuSO4 + H2O

It's already balanced, I want to work out the theoretical yield of this reaction to produce copper sulfate (and water). Please write out the equation so I can look it over and better understand this topic. I've already looked online, but the way it is formatted on the websites makes it hard to read. From what little I was able to make out I found sulfuric acid to be the limiting reactant. 

I will use 1g CuO and 20cm^3 sulfuric acid (0.5m) 
I sadly don't know the mole ratio, but i think you work it out something like
CuO
1g x 1 / 79.545 

Since 79.545  is the molar mass of 1g of CuO, or at least i think it is. I've re-upped the question with more values, realistically i wouldn't have minded if you inputted your own values since i just want to learn how to work it out.

Thanks.

Roger the Mole · Accepted Answer

You find the mole ratio by looking at the balanced equation.  In this case all the mole ratios are 1:1.

(1 g CuO) / (79.5457 g CuO/mol) = 0.0126 mol CuO
Supposing you mean 0.5 M H2SO4.  ("M" and "m" are different things.)
(0.020 L) x (0.5 mol/L) = 0.01 mol H2SO4

0.01 mole of  H2SO4 would react completely with 0.01 mole of CuO, but  there is more CuO present than that, so CuO is in excess and H2SO4 is the limiting reactant.

(0.01 mol H2SO4) x (1 mol CuSO4 / 1 mol H2SO4) = 0.01 mol CuSO4

Or if you prefer mass:
(0.01 mol CuSO4) x (159.6095 g CuSO4/mol) = 1.6 g CuSO4

Willard Thomas · Answer

CuO   +  H2SO4  =  CuSO4  + H2O

1.0g    +     36.8g   =     ?g       +   ?g

20cm^3 H2SO4 * [1.84g H2SO4 / 1 cm^3 H2SO4] = 36.8g  H2SO4

1.0g  CuO * [1 mol CuO / 79.546g CuO] = 0.0126 mol CuO

36.8g  H2SO4 * [1 mol H2SO4 / 98.079g H2SO4] = 0.3752 mol H2SO4

0.0126 mol CuO * [ 1 mol H2SO4 needed / 1 mol CuO] = 0.0126 mol H2SO4 are needed to react with the 0.0126 mol of CuO.

0.0126 mol  H2SO4 *[ 98.079g  H2SO4/ 1 mol H2SO4] = 1.236g  H2SO4

NOTE: Comparing the moles of the H2SO4 sample needed (0.0126 mol H2SO4) to that which is actually available (0.3752 mol H2SO4) we find therefore that:

[0.3752 mol H2SO4 – 0.0126 mol H2SO4] = 0.3636 mol H2SO4 in excess that remains after the chemical reaction runs its course.

0.3636 mol H2SO4 * [ 98.079g H2SO4 / 1 mol H2SO4] = 35.66g  H2SO4 of the sample remain after the completion of this reaction.

The theoretical yields for the compounds of CuSO4 and H2O are determined to be respectively as follows:

0.0126 mol H2SO4 * [1 mol CuSO4 / 1 mol H2SO4] *[159.610g CuSO4 / 1mol CuSO4] = 2.011g CuSO4.

0.0126 mol H2SO4 * [1 mol H2O / 1 mol H2SO4] * [18.015g H2O/1 mol H2O]
= 0.2270g H2O.

% yield of the reaction is  = [0.0126 mol H2SO4 /  0.3752 mol H2SO4]  x 100
% yield of the reaction is = 3.358% = approx..  3.36%
                                         = [ 1.236g H2SO4 / 36.8g H2SO4] x 100
                                         = 3.359% = approx..  3.36%
For additional assistance and information, please see also the following data link(s):

It is my sincere hope, and prayer that the data will be of a good benefit as they have been to me. “May God continue to richly bless you in all of your endeavors!”

Balancing Chemical Equations:
http://chemistry.about.com/cs/stoichiometry/a/aa042903a.htm
http://en.wikipedia.org/wiki/Chemical_equations
http://en.wikipedia.org/wiki/Stoichiometry
http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Stoichiometry_and_Balancing_Reactions
http://www.shodor.org/unchem-old/basic/stoic/index.html

Types of Chemical Reactions:
http://misterguch.brinkster.net/6typesofchemicalrxn.html
http://www2.ucdsb.on.ca/tiss/stretton/CHEM1/stoich2.html
http://crescentok.com/staff/jaskew/isr/chemistry/class11.htm
http://www.ric.edu/faculty/ptiskus/reactions/

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What is the theoretical yield of Copper Oxide with Sulfuric Acid?

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