3. Consider the following reaction, Fe2O3 + 3 CO → 2 Fe + 3 CO2
If 78.3g of Fe2O3 is available to react with 65.8g of CO;
How many grams of Iron will be produced?
• What is the limiting reagent?
How much excess reactant will be left over?
Solved
Show answers
More tips
- S Style and Beauty Is Photoepilation the Solution to Unwanted Hair Forever?...
- O Other What is a Disk Emulsifier and How Does it Work?...
- F Family and Home What does a newborn need?...
- F Family and Home Choosing the Right Car Seat for Your Child: Tips and Recommendations...
- F Food and Cooking How to Get Reconfirmation of Registration?...
- C Computers and Internet How to Get Rid of Spam in ICQ?...
- A Art and Culture Who Said The Less We Love a Woman, the More She Likes Us ?...
- F Family and Home How to Get Rid of Your Neighbors?...
- S Society and Politics How Could Nobody Know About the Dead Mountaineers?...
- H Health and Medicine How to Cure Adenoids?...
Answers on questions: Chemistry
- C Chemistry The function of beta-galactosidase is (science)...
- C Chemistry What is a safety rule while using glass beakers...
- C Chemistry Someone help please .. pleasaeeebeubdnficbc...
- C Chemistry Number 4. In the picture...
- A Arts Part A What is the author s point of view toward Azy s vocabulary mistake in The Think Tank ? The mistake proves that orangutans resist learning. The mistake suggests...
- P Physics What statement about gravity is true?...
- C Computers and Technology I need a best friend who is hot cute funny caring...
- B Biology Sports drinks are often recommended to prevent or treat dehydration. What kind of solution (isotonic, hypotonic, or hypertonic) do sports drinks provide outside...
- S Social Studies 1. What is Brown Blood disease in fish 2. What is the cause of brown blood disease...
- M Mathematics Please help me and I’ll give you brainiest...
Ответ:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M