# Stoichiometry assistance

1. Feb 22, 2012

1. The problem statement, all variables and given/known data

Nitric Acid is produced via the following equation:

$3NO_2 + H_2O \rightarrow 2HNO_3 + NO$

How many grams of $NO_2$ are required to produce 7.50 grams of $HNO_3$?

2. Relevant equations

$HNO_3 = 138 amu$
$NO_2 = 126 amu$

3. The attempt at a solution

I think I have a solution to this, but formulating these equations does not click in my head properly. I see other people do it over and over, but it hasn't clicked yet.

$7.50g HNO_3 * \frac{138g NO_3}{126g HNO_3}=8.21g NO_2$

I'm getting mixed up by thinking that this may be a shortcut. Well actually, I'm mixed up on a lot of it, like why can you just divide $\frac{138g NO_3}{126g HNO_3}$? I can see how it works nicely and how handy it is, but it's not intuitive to me yet so I'm having a hard time processing it.

Also, should I be adding Moles to the equation above, like this

$7.50g HNO_3 * \frac{138g NO_3}{1 Mol NO_3}*\frac{1 Mol HNO_3}{126g HNO_3} =8.21g NO_2$

and then perform the arithmetic? However, if I do it that way I cannot cross out some of the components properly.

Any advice would be appreciated. I don't know why but Chemistry isn't clicking the way I was hoping it would. I'm not having nearly this much trouble in my calc class.

2. Feb 23, 2012

### Staff: Mentor

Your final result is correct (8.21 g it is), but this:

is incorrect.

Molar mass of HNO3 is 63 g/mol, molar mass of NO2 is 46 g/mol. Masses that you listed are multiplied by the stoichiometric coefficients which you shouldn't do. Stoichiometric coefficients are another conversion factor:

$$7.50g HNO_3 = 7.50 g HNO_3 \frac {1 mol HNO_3}{63 g HNO_3} \times \frac {3 moles NO_2} {2 moles HNO_3} \times \frac {46 g NO_2}{1 mol NO_2} = 8.21 g NO_2$$

First you convert 7.50 g HNO3 to number of moles, then you use stoichiometric coefficients to calculate number of moles of NO2, then you convert moles of NO2 to its mass.

What you did is similar to the method described here: http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

3. Feb 23, 2012

Thank you, Borek.

I knew I had made some shortcut that I wasn't accounting for, but I couldn't quite put my finger on it. I appreciate the assistance.

I also appreciate the link. The method shown there makes much more sense to me than this way of doing it. In fact, this is how I was beginning to figure things out on my own, but all resources that I could locate showed how to do it the way listed in this problem.

I wonder why that method isn't more widely taught? I'll have to digest it for a while, but it does seem more intuitive.

Thanks again,
Mac

4. Feb 23, 2012

### Staff: Mentor

No idea. That's the way I was taught to do stoichiometry problems back in seventies in my neck of the woods. Few years ago I asked American chemistry teachers for comments on the content of the page and I was told that students have no idea why they should use "different molar masses" for different reactions. It is a misunderstanding of the method and I don't buy it.

Glad that you find it intuitive.