# chemistry/math solution stoichiometry

• Nov 17th 2008, 05:58 PM
daydreembelievr
chemistry/math solution stoichiometry
A 9.3 gram sample of chromium (III) nitrate is placed in 30.0 mL of a 0.100 M (molarity) barium hydroxide solution. What is the molarity of the chromium (III) nitrate solution before the reaction occurred?

I found the balanced equation to be:
2Cr(NO3)3 + 3BaOH2 --> 2CrOH3 + 3Ba(NO3)2

After I do this am I suppose to convert 9.3 grams of chromium (III) nitrate into either volume or molarity so I can use the formula:
V1xM1=V2xM2 ?????

If not, what should my next step be? Thanks so much for any help. :D
• Nov 17th 2008, 06:09 PM
o_O
$2\text{Cr(NO}_3)_3 + 3\text{Ba(OH)}_2 \ \rightarrow \ 2\text{Cr(OH)}_3 + 3\text{Ba(NO}_3)_2$

Actually you don't need $M_1V_1 = M_2V_2$. We don't even need the reaction because we don't even consider it.

You want the molarity (i.e. # of moles divided by volume) of chromium (III) nitrate. We're given that they're 9.3 grams of it (which can be converted into moles) and we're given that it's put into a 30 mL solution.

So to find its molarity ... ?
• Nov 17th 2008, 06:15 PM
daydreembelievr
When I find the molar mass do I consider the 2 in front of the Cr and the subscript 3 of the nitrate? Thanks :D
• Nov 17th 2008, 06:20 PM
o_O
No for the first question, yes for the second. The 2 in front of $\text{Cr(NO}_3)_3$ means that 2 of chromium (III) nitrate is needed for it to react with the barium hydroxide. It has nothing to do with its molar mass, only the subscripts of the compound give you that information.