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Chemical Equation Help: ASAP! Please.

Posted: Sun Dec 31, 2006 12:55 pm
by CreationArtist
Please be kind enough to help out.
I really need this help ASAP.

HBr + Mg(OH)2 -> H2O + NaCl

H2S + Ba(OH)2 -> H2O + NaCl


I don't know if these are correct/balanced.
Can you help me out ASAP?

Thanks!

Also, how do you do this problem?

4AU + 8KCN + O2 + 2H2O -> 4KAu(CN)2 + 4KOH

How many moles of KCN are required to extract 29.0 g pf gold (about an ounce) of gold?


Also, the other day I could not remember how to do this type of problem:

Nitrogen oxide reacts instantly with oxygen gas to form nitrogen dioxide. Write the equation. In one experiement, 0.866 moles of NO is mixed with 0.503 moles of oxygen gas. Calculate which of the reactants is the limiting one? How many moles of nitrogen dioxide are produced?

or this simple one that I simply forgot how to use 6.02EE23.
The carat is the unit of mass used by jewelers. One carat is exactly 200 mg. How many carbon atoms are present in a 24-carat diamond?


If you could walk me through each problem, that would be greatly appreciated. Thanks.

And sorry for all of the questions (my midterm is coming up and I have a lot of studying.. we don't use that great of a book either) and I really need to know how to do all of these very shortly.

Re: Chemical Equation Help: ASAP! Please.

Posted: Mon Jan 01, 2007 12:24 pm
by chempass
CreationArtist wrote: Please be kind enough to help out.
I really need this help ASAP.

HBr + Mg(OH)2 -> H2O + NaCl

H2S + Ba(OH)2 -> H2O + NaCl

I don't know if these are correct/balanced.
Can you help me out ASAP?

Thanks!
They are both not correct.
Here are balanced and correct equations:
2HBr + Mg(OH)2 = MgBr2 + 2H2O
H2S + Ba(OH)2 = BaS + 2H2O
CreationArtist wrote: Also, how do you do this problem?

4Au + 8KCN + O2 + 2H2O -> 4KAu(CN)2 + 4KOH

How many moles of KCN are required to extract 29.0 g pf gold (about an ounce) of gold?
m(Au) = 29.0 g (given)
M(Au) = 196.97 (atomic weight of Au: http://www.webqc.org/periodictable-Gold-Au.html)
n(Au) = m(Au) / M(Au) (calculate how many moles of Au are in 29.0 g)
n(KCN) = (n(Au) / 4) * 8 (calculate how many moles of KCN are required)
CreationArtist wrote: Also, the other day I could not remember how to do this type of problem:

Nitrogen oxide reacts instantly with oxygen gas to form nitrogen dioxide. Write the equation. In one experiement, 0.866 moles of NO is mixed with 0.503 moles of oxygen gas. Calculate which of the reactants is the limiting one? How many moles of nitrogen dioxide are produced?
2NO + O2 = 2NO2

0.866/2 < 0.503 therefore NO is limiting
n(NO2) = n(NO) * 2 /2 = 0.866
CreationArtist wrote: or this simple one that I simply forgot how to use 6.02EE23.
The carat is the unit of mass used by jewelers. One carat is exactly 200 mg. How many carbon atoms are present in a 24-carat diamond?
m(C) = 24 * 200 mg (calculate the mass of carbon)
n(C) = m(C) / M(C) (calculate number of moles of carbon)
N(C) = n(C) * 6.02E23 (calculate number of carbon atoms)

Posted: Mon Jan 01, 2007 1:24 pm
by CreationArtist
Thank you so much, I PM'd you back.

Posted: Fri Jan 05, 2007 6:59 pm
by chempass
CreationArtist wrote: Just making sure I understood the two that you didn't make the calculations on.. I got:

.0046 moles of KCN

and for the other I got:

2.4EE26 Atoms
1) 29.0 / 196.97 * 2 =0.294 moles of KCN

2) 24 * (200 / 1000) / 12.01 * 6.02E23 = 2.419E+23 atoms
CreationArtist wrote:
And sorry for all of the hassle (it's much appreciated), but it would really help if you could quickly show me how to do this:

Platinum forms two different compounds with chlorine. One contains 26.7% Cl by mass. The other contains 42.1% Cl by mass. Determine the empirical formulae of the two compounds.

100 - 26.7= 73.3

26.7% of Cl (MM=35.453)
73.3% of Pt (MM=195.08)

Cl
26.7g. x 1 mole/35.453 = 0.753 moles

Pt
73.3g. x 1 mole/195.08 = 0.376 moles

0.753/0.376 = 2

0.376/0.376 = 1

2Cl : 1Pt ratio, therefore the first compound is:

PtCl2

and you do the same thing for the other compound?
Yes, you need to do the same thing. (you will get PtCl4 )

It's better to post here rather than PM.