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.

 

[DeeNos]

I can understand your struggles with stoichiometry and balancing equations. It can be a challenging concept to grasp at first, but with practice and understanding of the fundamental principles, it will become easier.

First, let's break down the equation and understand what it is telling us. In this reaction, 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3 and 1 mole of NO. This can be represented as a ratio: 3 moles NO2 : 1 mole H2O : 2 moles HNO3 : 1 mole NO.

Now, let's look at the given information. We are given the mass of HNO3, which is 7.50 grams. We can convert this to moles by using the molar mass of HNO3 (138 g/mol): 7.50 g HNO3 * (1 mol HNO3 / 138 g HNO3) = 0.0543 mol HNO3.

Next, we can use the mole ratio from the balanced equation to determine the moles of NO2 needed to produce this amount of HNO3. Since the ratio is 3 moles NO2 : 2 moles HNO3, we can set up a proportion: 3 moles NO2 / 2 moles HNO3 = x moles NO2 / 0.0543 moles HNO3. Solving for x gives us x = 0.0815 moles NO2.

Finally, we can convert the moles of NO2 to grams by using the molar mass of NO2 (46 g/mol): 0.0815 mol NO2 * (46 g NO2 / 1 mol NO2) = 3.75 g NO2.

So, 3.75 grams of NO2 are required to produce 7.50 grams of HNO3 in this reaction.

To answer your question about why we can divide the molar masses in the equation, it is because of the concept of molar ratios. In a balanced equation, the coefficients represent the number of moles of each substance involved in the reaction. So, by dividing the molar masses, we are essentially converting from grams to moles and using the coefficients as the conversion factor. This is why we can cross out the units (g HNO3 and g NO2