Need to write a balanced chemical equation for the reaction between hydrogen selenide and tin. I know that the hydrogen selenide decomposes producing solid tin selenide and hydrogen gas but how would I balance that in the equation. Any help gratefully received.
I can write and balance an equation for H2Se and Sn but I am not convinced one of the products is H2 gas. How do you know H2 gas is one of the products. Do you have any indication of the other product besides H2 gas? If this is a problem, please post the entire problem.
It says that the composition of hydrogen selenide can be determined by heating tin in a measured volume of gas. The hydrogen selenide is decomposed, producing solid tin selenide and hydrogen gas. If the temperature and pressure are unchanged, the gas volume when the decomposition is over is the same as it was to begin with. I need a balanced equation for the reaction between hydrogen selenide and tin, and explain why the equation accounts for the fact that the gas volume is unchanged. So far I have H2Se + Sn = SnSe3 + H2, not sure if this is correct, thanks for replying so quick.
The two common oxidation states of Sn are +2 and +4. If it forms tin(IV) selenide (stannic selenide), the reaction is
2H2Se + Sn ==>2H2 + SnSe2
If it forms tin(II) selenide (stannous selenide), the reaction is
H2Se + Sn ==>H2 + SnSe
I simply don't know enough about the chemistry of Se to know which it does; however, in either case the volume will not change. In the first equation, 2 mols H2Se gas produce 2 mols H2 gas. In the second equation, 1 mol H2Se gas produces 1 mol H2 gas. If your problem lists "producing tin selenide" tin(II) or tin(IV) you will know which equation to use. There is no evidence I can find that would account for SnSe3.
My educated guess, from sulfur chemistry, which is in the same group as Se, I think SnSe will be formed.
I hope this helps but check it out.
Many thanks for your time
WOW....alot of big words there!
2H2Se (g) + Sn (s) = Sn Se2 (s) + 2H2(g)
The composition of hydrogen selenide can be determined by heating tin (Sn) in a measured volume of the gas. The hydrpgen selenide is decomposed, producing solid tin selenide (SnSe) and hydrogen gas.
2H2Se+2Sn - 2SnSe+4H
Dr Bob222, this is the third different question you have really helped me to understand so far, you rule man!
I take it ur doing S103 then
It seems like you have already received some help, but I'd be happy to explain and confirm the balanced chemical equation for the reaction between hydrogen selenide (H2Se) and tin (Sn).
The correct balanced equation is:
2H2Se (g) + Sn (s) → SnSe2 (s) + 2H2 (g)
The reaction shows that two molecules of hydrogen selenide gas react with one molecule of solid tin to produce one molecule of tin selenide and two molecules of hydrogen gas.
To balance this equation, you need to ensure that the same number of atoms of each element is present on both sides of the reaction arrow. Here's an explanation of how to balance this equation:
1. Start by counting the atoms of each element present on both sides of the equation:
- Hydrogen (H): 2 on the left, 2 on the right
- Selenium (Se): 2 on the left, 2 on the right
- Tin (Sn): 1 on the left, 1 on the right
2. Balance each element one at a time:
- Balance hydrogen (H) atoms by adding a coefficient of 2 in front of H2 on the right-hand side.
- This gives us 2H2Se (g) + Sn (s) → SnSe2 (s) + 2H2 (g)
- Check that the number of hydrogen atoms is balanced:
- 2 on the left, 4 on the right (2x2)
- Balance selenium (Se) atoms by adding a coefficient of 2 in front of SnSe2 on the right-hand side.
- This gives us 2H2Se (g) + Sn (s) → 2SnSe2 (s) + 2H2 (g)
- Check that the number of selenium atoms is balanced:
- 2 on the left, 4 on the right (2x2)
- Tin (Sn) atoms are already balanced, so we don't need to make any changes.
- Check that the number of tin atoms is balanced:
- 1 on the left, 2 on the right (2x1)
By following these steps and making the appropriate changes, you can balance the equation correctly.
I hope this explanation helps! Let me know if you have any further questions.