Stoichiometry Mole Conversion Problems

[Cereal]

Hey there, I've been trying for the past 45 minutes to work on my online homework but I have made hardly any progress. Here are the 2 problems I'm having trouble with:



1- Calculate the number of molecules of NBr3 in 37.5 g NBr3. Use a molar mass with at least as many significant figures as the data given.

2-CH4(g) + Cl2(g) --> CHCl3(l) + HCl(g)

What mass of HCl can be produced from 1.60 g of methane and 10.0 g of chlorine in the above unbalanced reaction? Use at least as many significant figures in your molar masses as in the data given.

The problem in the first one is that I'm going from mass to molecules. I've only done problems involving moles and mass. Any hints as to how to get started?

For the second one, I think I balanced the equation right. 2CH4 + 6Cl2 --> 2CHCl3 + 6HCl? I also see the problem is converting mass to mass. However, with methane AND chlorine involved, I'm lost as to where to go from here.

All help is appreciated, thanks!

 

[AngelShare]

*Groans* I'm doing exactly that in my online course.:yuck:

In case this helps you (It didn't help me but I'm convinced I'm just slow...:tongue: ), here is what we were given...

http://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Stoichiometry.html

http://www.chem4kids.com/files/react_stoichio.html

I'll have to look over it again myself.

 

[Blueshift5]

I understand that stoichiometry and mole conversions can be challenging concepts to grasp. It is important to have a solid understanding of these concepts in order to successfully solve problems like the ones you are struggling with.

For the first problem, you are correct that you need to convert from mass to molecules. To do this, you will need to use the molar mass of NBr3, which is 187.71 g/mol. This tells us that for every 187.71 g of NBr3, there are 6.022 x 10^23 molecules. Using this conversion factor, you can set up a dimensional analysis problem to solve for the number of molecules in 37.5 g of NBr3. Remember to always check your units and make sure they cancel out to give you the correct final unit.

For the second problem, you have correctly balanced the equation. To solve for the mass of HCl produced, you will need to use the molar masses of each reactant and product. This tells us that for every 16.04 g of CH4, there are 36.46 g of HCl produced. Similarly, for every 70.91 g of Cl2, there are 36.46 g of HCl produced. Using these conversion factors, you can set up a dimensional analysis problem to solve for the mass of HCl produced. Again, make sure to check your units and use the correct number of significant figures in your final answer.

I hope this helps you get started on solving these problems. Remember to always double check your work and use the appropriate conversion factors. If you continue to struggle, don't hesitate to seek out additional resources or ask for help from your instructor or peers. Good luck!