I'm taking chemistry 1 in high school and we're doing something called stoichiometry, and if you know chemistry you should know what this is. Basically, the teacher gave us a load of problems to do and told us to figure out the equations and we also learned that prediction stuff but not in depth. Here's an example,
1. Lightning discharges in the atmosphere catalyze the conversion of nitrogen to nitric oxide. How many grams of nitrogen would be required to make 25 grams of nitric oxide this way?

2. Are 15 g of chlorine gas enough to replace 30 g of of broine from a solution of sodium bromide?

I think they're both gram - gram equations. Help! thanks

miNi

Korea (South)2010 Posts

February 25 2008 04:45 GMT

You need to write the chemical equations, then use dimensional analysis
1. N2 + O2 -> 2NO
25g NO * (1 mol NO / 30.1g NO) * (1 mol N2 / 2 mol NO) * (28.02g / 1mol N2)

2. You need to see whether 15g is more moles than 30g Br
Convert them to moles, and see whether it's true

GrayArea

United States872 Posts

February 25 2008 04:47 GMT

I agree with the guy above me.

miNi

Korea (South)2010 Posts

February 25 2008 04:49 GMT

nitrogen and oxygen are diatomic molecules, so the mole ratios will not be 1 to 1

Steelflight-Rx

United States1389 Posts

February 25 2008 04:50 GMT

okay so yea i only read the first one its a gram to gram problem. You need to figure out the formula of nitric oxide to solve the problem, which happens to be NO.

1.) 25 g nitric oxide times 1/molar mass nitric oxide <-- this the grams cancel, giving u moles of nitrogen

2.) moles of nitric oxide (from step one) times 1 mole N / 1 moles NO <-- this step, you cancel out the nitric oxide giving you just moles of nitrogen.

3.) moles of N times molar mass of nitrogen (grams per mole) <-- this step the moles cancel out, leaving you with the mass of nitrogen needed.

took me a while to write this i bet sum1 already answered it

Plutonium

United States2217 Posts

February 25 2008 04:51 GMT

For the first one, Nitrogen is in the form of Nitrogen Gas, which has the chemical formula N2 and a mass of 28.02 g/mol

GrayArea

United States872 Posts

February 25 2008 04:51 GMT

On February 25 2008 13:45 miNi wrote:
You need to write the chemical equations, then use dimensional analysis
1. N2 + O2 -> 2NO
25g NO * (1 mol NO / 30.1g NO) * (1 mol N2 / 2 mol NO) * (1 mol N2 / 28.02 g)

2. You need to see whether 15g is more moles than 30g Br
Convert them to moles, and see whether it's true

The bold part is wrong, it should be flipped upside down.

Steelflight-Rx

United States1389 Posts

February 25 2008 04:52 GMT

to mini: i dont think the fact that they are diatomic matters for this problem, its assuming 1 mole of nitrogen yields one mole of NO

miNi

Korea (South)2010 Posts

February 25 2008 04:54 GMT

ty grayarea,
steel: nitrogen exists in the atmosphere as N2, so yes i believe that it does matter

Steelflight-Rx

United States1389 Posts

February 25 2008 04:57 GMT

#10

yeah both our equations give the same answer but i guess for the sake of writing a correct chemical equation, yeah its N2. I misread what u were saying i thought u meant N2 was going to give a different answer than just N or something =P

IefNaij

Canada291 Posts

February 25 2008 04:59 GMT

#11

N2 + O2 -> 2NO assuming Nitrogen gas is the limiting reactant, since its not stated

n(NO) = mass/Molar mass
= 25/30
= 0.83 mol

since mol ratio is 1:2

n(N2) = 0.415mol
m(N2) = mol * molarmass
= 0.415 * 28
= 11.6g

my first year in high school chem, so im not 100% sure

GrayArea

United States872 Posts

February 25 2008 05:04 GMT

#12

On February 25 2008 13:59 IefNaij wrote:
N2 + O2 -> 2NO assuming Nitrogen gas is the limiting reactant, since its not stated

n(NO) = mass/Molar mass
= 25/30
= 0.83 mol

since mol ratio is 1:2

n(N2) = 0.415mol
m(N2) = mol * molarmass
= 0.415 * 28
= 11.6g

my first year in high school chem, so im not 100% sure

Well, technically oxygen would be the limiting reactant because air is composed of about 70% nitrogen and 20% oxygen. However, in this case, it is irrelevant to designate the limiting reactant because it is being performed in the atmosphere with such an abundance of both gases. I didn't do any of the calculations so I don't know whether the rest of your post is correct or not.

il0seonpurpose

Korea (South)5638 Posts

February 25 2008 05:22 GMT

#13

Ok, thanks for all the details! All I needed was help on the reaction prediction, my teacher never taught some of the stuff you all are talking about.