Determine the mass of sodium chloride produced when 2.5 moles of chlorine ga

Click For Summary
SUMMARY

The discussion focuses on calculating the mass of sodium chloride (NaCl) produced from the reaction of 2.5 moles of chlorine gas (Cl2) with sodium (Na). The balanced chemical equation is 2 Na + Cl2 = 2 NaCl, indicating a stoichiometric ratio of 2:1. Given that chlorine is the limiting reagent, the calculation shows that 5 moles of NaCl are produced. The final mass is calculated using the formula mass = moles x molar mass, resulting in 292.2 grams of NaCl.

PREREQUISITES
  • Understanding of stoichiometry and chemical equations
  • Knowledge of molar mass calculations
  • Familiarity with limiting reagents in chemical reactions
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study stoichiometry in chemical reactions
  • Learn about limiting and excess reagents in reactions
  • Practice calculating molar masses of compounds
  • Explore more complex chemical reaction problems
USEFUL FOR

Chemistry students, educators, and anyone preparing for exams involving chemical reactions and stoichiometry.

BittersweetLove
Messages
19
Reaction score
0
[SOLVED] Determine the mass of sodium chloride produced when 2.5 moles of chlorine ga

Okay...I know how to solve half the problem. I know how to balance the problem but I'm not sure what to do next. I'm really bad at this stuff.

Here's the question:

Determine the mass of sodium chloride produced when 2.5 moles of chlorine gas reacts with sodium. Na + Cl2 NaCl (be sure to balance the reaction first!)

And this is the balanced equation:

2 Na + Cl2 = 2 NaCl

Anyone have any suggestions?
 
Physics news on Phys.org
Assuming it is reacting with excess sodium, Chlorine becomes the limiting reagent.
So the moles of NaCl should be 5 moles (since 2 : 1 = 2 stoichiometric ratio).
Using the formula:
#of moles = mass(g)/molar mass
Some basic algebra:
Mass = 5 x (22.99 + 35.45)
= 292.2 g

Hmmm...doesn't look right..
Did something wrong, anyone correct me please?
 
Looks fine.

Keep in mind for the future that it is better to guide the OP through suggestions, questions and hints than to write up the complete solution. The former approach is often a much more effective teaching tool. And if you haven't already, please take a look at our posting Guidelines.
 
Thanks. Hopefully I'm remember that for the test. *fingers crossed* It seems easy enough but when you suck at this stuff, it's hard to remember everything. Thanks again. 8)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
12K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K