itzandrea456
07.05.2020 •
Chemistry
Suppose 0.950 L of 0.410 MH,SO, is mixed with 0.900 L of 0.240 M KOH. What concentration of sulfuric acid remains
after neutralization?
Solved
Show answers
More tips
- F Food and Cooking Ginger: A Universal Ingredient for Various Dishes...
- A Auto and Moto Which alarm system to choose?...
- C Computers and Internet Porn Banner: What It Is and How to Get Rid Of It?...
- F Food and Cooking How many stages of coffee roasting are there?...
- F Food and Cooking From Latte to Espresso: Which Coffee Drink is the Most Popular on Earth?...
- F Food and Cooking Experts Name Top 5 Healthiest Teas...
- C Computers and Internet Google Search Tips and Tricks: Everything You Need to Know...
- H Health and Medicine Discover the Hidden Principles and Real Results of the Japanese Diet...
- H Health and Medicine Liver Cleansing - Rejuvenation for the Body?...
- S Style and Beauty Is Photoepilation the Solution to Unwanted Hair Forever?...
Answers on questions: Chemistry
- M Mathematics The FDA is testing a new ointment designed to deliver 3.5 micrograms of active ingredient to each square centimeter of skin. Nine(9) subjects apply the ointment and then...
- H History Market research can be expensive. Which of the following is the best explanation for why a businessperson would be willing to invest in such research? A. To raise prices...
- B Biology When is formed when atoms of one type react with atoms of other types...
- M Mathematics There are bots sending links to sus site dont click on it it will probably steal ur cookies...
- M Mathematics Svaries directly as t. if s is 20 when t is 4, then t is a0 when s is 30...
Ответ:
The remaining concentration of H2SO4 is 0.152 M
Explanation:
Step 1: Data given
Volume of H2SO4 = 0.950 L
Molarity H2SO4 = 0.410 M
Volume of KOH = 0.900 L
Molarity of KOH = 0.240 M
Step 2: The balanced equation
H2SO4 + 2KOH → K2SO4 + 2H2O
Step 3: Calculate moles
Moles = molarity * volume
Moles H2SO4 = 0.410 M * 0.950 L
Moles H2SO4 = 0.3895 moles
Moles KOH = 0.240 M * 0.900L
Moles KOH = 0.216 moles
Step 4: Calculate the limiting reactant
For 1 mol H2SO4 we need 2 moles KOH to produce 1 mol K2SO4 and 2 moles H2O
The limiting reactant is KOH. It will completely be consumed (0.216 moles).
H2SO4 is in excess. There will react 0.216/2 = 0.108 moles. There will remain 0.3895 moles - 0.108 moles = 0.2815 moles
Step 5: Calculate the concentration of H2SO4 remaining
[H2SO4] = moles / volume
[H2SO4] = 0.2815 moles / 1.85 L
[H2SO4]= 0.152 M
The remaining concentration of H2SO4 is 0.152 M
Ответ: