## hlilly2413 one year ago How many grams of aluminum nitrate (Al(NO3)3 are needed to prepare 250mL of a solution that has a concentration of 0.200 moles of NO3- per liter of solution? The molar mass of Al(NO3)3 is 213.01 g/mole, and there are 3 moles of NO3- per mole of Al(NO3)3. *All 3s in chemical formula are subscripts. How in the world do I go about this? Please help

1. anonymous

Your guiding thread will be how well you are at figuring out the units of everything and keeping an eye on where you want to go and what you're given. Best thing to do is write everything out as well as you can, and until you've had some experience this is a bit tricky! So let's identify what we're actually given: $250 \ mL$ $0.200 \ \frac{mol \ (NO_3)^-}{L }$ $\frac{1 \ mol \ Al (NO_3)_3}{3 \ mol \ (NO_3)^-}$ $213.01 \frac{g \ Al(NO_3)_3}{mol \ Al (NO_3)_3}$ Spend some time making sure this makes sense, this is really the hardest part. Once you've determined all the units to go with the values, the rest will fall into place. We now recognize that we need grams of $$Al(NO_3)_3$$ and we see this on that last value above, so we need to remove that value of moles of $$Al (NO_3)_3$$ on the bottom. So in order to do that, we multiply by something that has moles of $$Al (NO_3)_3$$ in the top, so we look and sure enough we have: $$\frac{1 \ mol \ Al (NO_3)_3}{3 \ mol \ (NO_3)^-}$$ which has it in the top, so we multiply them, and they cancel out since anything divided by itself is just 1, even words in conversions, like this! $\left(213.01 \frac{g \ Al(NO_3)_3}{mol \ Al (NO_3)_3} \right) \left(\frac{1 \ mol \ Al (NO_3)_3}{3 \ mol \ (NO_3)^-}\right)$ $\left(213.01 \frac{g \ Al(NO_3)_3}{\cancel{mol \ Al (NO_3)_3}} \right) \left(\frac{1 \cancel{\ mol \ Al (NO_3)_3}}{3 \ mol \ (NO_3)^-}\right)$ $213.01*\frac{1 }{3 } \frac{g \ Al(NO_3)_3}{\ mol \ (NO_3)^-}$ So now we have a new thing in the denominator to cancel, but still keep our 1/3 since only the units went away not the number itself. Try to find the next one that will cancel out the $$mol \ (NO_3)^-$$ in the denominator, remember we want to hold onto grams of Aluminum Nitrate in the end, so if we're lucky everything will perfectly cancel out except for this in the end. Give it a try doing the next step and I'll help. if you need it.

2. hlilly2413

Hey @Woodward thank you for your assistance with beginning this problem. Ok, so the next thing I was thinking about doing was tacking the following on to what you already started. $\frac{ 1 mol (NO _{3)-} }{ .200 mol (NO _{3)-} }$ However, because I figured that would cancel out the 1 mol and the 3 mol, but normally I cancel it out the other way so...is that the next step or should I be doing something else instead?

3. hlilly2413

However, because doesn't make any sense. Ignore both of those words in that sentence.

4. hlilly2413

Wait that might not be right because should it be 250mL/.200 moles (N03)- ?

5. anonymous

Yeah, good you're headed in the right direction, so for right now you have $0.200 \frac{ mol \ (NO_3)^-}{L}$ Multiplying this by what you have will cancel out the moles of nitrate ions, and leave you with Liters. Also you correctly recognized that we have to convert next, so we gotta turn L into mL to allow us to deal with it, but I think you might have used the wrong conversion factor, at any rate I'll go ahead and put it: $1 \ L = 1000 \ mL$

6. hlilly2413

I used the 250mL because that was a given in the problem. Ok, I'm going to show you what I have written on my paper to make sure it's correct. The number comes out sort of large which concerns me, so...it make take a sec for me to type it. I'm new at using the equation section on here.

7. hlilly2413

$213.01 \frac{ g }{ mol } Al(NO _{3)3} \times \frac{ 1 mol Al(NO _{3)3} }{ 3 mol (NO _{3)-} }\times \frac{ 1 L }{.200 mol (NO _{3)-} }$

8. hlilly2413

the $\frac{ 1 L }{ .200 mol (NO _{3)-} } \times \frac{ 250 mL of soln}{ 1 L }$ but to me it doesn't look right so

9. anonymous

$213.01 \frac{ g }{ mol } Al(NO _3)_3 \times \frac{ 1 mol Al(NO _3)_3 }{ 3 mol (NO _3)^-}\times \color{red}{\frac{0.200 \ mol \ (NO_3)^-}{L}} \times \frac{ 1 L }{1000 \ mL}$ $\times 250 \ mL$

10. anonymous

Make sure everything cancels I think that works out, it's just so long I couldn't fit it all on one line haha

11. hlilly2413

Ok! I was reworking it in my notebook and noticed that I did have the inverse of .200M instead of the correct way. I also forgot to add the 1L/1000mL part. I wasn't sure if I could just multiply by 250mL at the end or if I needed something else. So, the answer came out to be 3.55 g of $Al(NO _{3)3}$

12. hlilly2413

Whew. Thank you so much. I really appreciate it. Out of 135 chem probs for HW I had 3 that I couldn't figure out and of course they were all from the same 11 question packet. Odds were not in my favor. Now I only have one more to do! Thanks again for helping me walk through it @Woodward

13. hlilly2413

helping walk me through it* my brain is mush. :)

14. anonymous

Yeah! No problem, there's a lot happening here and it's easy to get lost on this one. Glad I could help you clear it up haha