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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Cuddlekins on March 12, 2011, 08:18:43 pm

Title: Refresh my memory please!
Post by: Cuddlekins on March 12, 2011, 08:18:43 pm

a bit rusty on chem, just to clarify.
Say a chemical reaction is 70% efficient and the question is asking for the mass of a substance. Would you multiply your mass of the substance answer with 70/100 ?

thanks

Title: Re: efficiency of chemical reactions
Post by: thushan on March 12, 2011, 09:03:03 pm

Yup.

Title: Re: efficiency of chemical reactions
Post by: Cuddlekins on March 13, 2011, 12:22:03 pm

If the process below is only 60% efficient, determine the mass of C3H5N3O9 required to produce
25 g of CO2.

n(CO2)= 0.568...

n(C3H5N3O9)= 0.568.. * 4/12
=0.189...

mass(C3H5N3O9)= 0.189... * 227
=42.99. * 0.60

i ended up getting 25.794g,but this isnt correct :S
Can someone check my working out,

Thanks heaps

Title: Re: efficiency of chemical reactions
Post by: thushan on March 13, 2011, 02:10:57 pm

Working is fine, except last step:

you want 42.99/0.60 (or 42.99 x 100/60) as we know that we need a higher than normal amount of C3H5N3O9 to produce the same amount of CO2.

Title: Re: Refresh my memory please!
Post by: Cuddlekins on March 13, 2011, 10:07:09 pm

     A 10.53 g sample containing a mixture of KCl and KClO3 is heated and 1.53 g of oxygen is liberated. Determine the % by weight of the KClO3 in the original mixture.

Thanks

Title: Re: Refresh my memory please!
Post by: Mao on March 13, 2011, 11:11:13 pm

The reaction is 2KClO3 --> 2KCl + 3O2

Using given data and this equation, you can find n(O2) --> n(KClO3) --> m(KClO3) --> %w(KClO3)

Title: Re: Refresh my memory please!
Post by: Cuddlekins on March 15, 2011, 03:27:48 pm

thanks i got it :)

Now this question
A mixture of Cu(NO3)2 and Cu(NO3)2.2.5H20 has a mass of 3.150 g. Upon heating the water is driven off and the mass of the Cu(NO3)2 is 2.853 g. Determine the % by weight of the Cu(NO3)2.2.5H20 in the original mixture.

I'm not sure how the equation will look like ?
i know first i calculated the mol of Cu(NO3)2.2.5H20, then im not sure about the mol ratio :S

Thanks

Title: Re: Refresh my memory please!
Post by: fady_22 on March 15, 2011, 03:36:20 pm

Here's the solution (I did that question in the ILT):

Title: Re: Refresh my memory please!
Post by: Cuddlekins on March 15, 2011, 08:19:45 pm

finding the mol of Cu(NO3)2.2.5H20. The dividing it by 2.5 was the problem, how did you get 2.5?

i thought Cu(NO3)2.2.5H20, means theres a 2 in front of the 5H20 :S:S

Title: Re: Refresh my memory please!
Post by: fady_22 on March 15, 2011, 08:56:31 pm

Think of it this way:
(when heat is applied)
Cu(NO3)2.2.5H20----->Cu(NO3)2+ 2.5H20

As you know the mole of H20, you can use stoich to find the number of mole, and thus mass, of Cu(NO3)2.2.5H20. Then you can deduce the percentage by mass of Cu(NO3)2.2.5H20 in the mixture (as the Cu(NO3)2 does not undergo a reaction by heating).