![j1theking18](/avatars/36751.jpg)
j1theking18
17.07.2019 •
Chemistry
What is the limiting reactant in the reaction of 12.0 g of so2 with 8.0 g of h2s and their reaction? what mass of sulfur will be produced?
Solved
Show answers
More tips
- A Animals and plants 10 Tips for Growing Delicious and High-Quality Tomatoes in Your Garden...
- H Health and Medicine Tick Traps: How to Remove Them Safely and Effectively...
- F Family and Home How to Keep Your Home Warm: Tips and Tricks...
- C Computers and Internet How Much Does an iPhone Cost in America?...
- S Style and Beauty Discover the Art of Nail Design: How Do You Paint Your Nails?...
- F Food and Cooking Discover How to Properly Prepare Dough for Rasstegai...
- F Family and Home Ways to Attract Wealth into Your Home...
- G Goods and services Stock center - a modern way of optimizing logistics...
- F Family and Home How to Teach Your Child to Read?...
- F Food and Cooking How to properly cook shrimp: tips from professionals...
Answers on questions: Chemistry
- E Engineering Alternating current on a power line oscillates at 60 Hz. Calculate the wavelength and determine whether transmission line effects are seen on a power line that...
- B Business The asset s book value is $70,000 on june 1, year 3. on that date, management determines that the asset s salvage value should be $5,000 rather than the original...
- M Mathematics Aliyah ate 2/5 of her lunch, and Chanel ate 1/3 of her lunch. Which girl ate more lunch?...
Ответ:
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
Ответ:
Your
3 x +5