Solving for NO2 in Nitric Acid Production Equation | Stoichiometry Assistance

[MacLaddy]

Homework Statement

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?

Homework Equations

HNO_3 = 138 amu
NO_2 = 126 amu

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.

 

[Borek]

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

HNO_3 = 138 amu

NO_2 = 126 amu

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

 

[MacLaddy]

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

 

[Borek]

I wonder why that method isn't more widely taught?

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.