When 79.1g of zinc are reacting with 1.05L of a 2.00 mol hydrochloric acid solution to produce zinc chloride and hydrogen gas, which reactant will be in excess and by how much? Calculate the number of grams of each product.
I want the solution and answer, ( If you could explain how to do it )
Thank you :)
If you could also answer this one
When 1.21 mol of zinc are reacting with 2.65 mol of hydrochloric acid to produce zinc chloride and hydrogen gas, which reactant will be in excess and by how much? Calculate the number of grams of each product.
Thanks a lot ><
Need help with Chemistry "Limiting Reagents".thank you?
2008-12-13 8:25 pm
回答 (2)
2008-12-13 8:36 pm
✔ 最佳答案
moles Zn = 79.1 g / 65.38 g/mol= 1.21moles HCl = 1.05 L x 2.00 M = 2.10
Zn + 2 HCl >> ZnCl2 + H2
the ratio between Zn and HCl is 1 : 2 so HCl is the limiting reactant ( 1.21 x 2 = 2.42 moles HCl needed)
the ratio between HCl and ZnCl2 is 2 : 1
moles ZnCl2 = 2.10 / 2 = 1.05
mass ZnCl2 = 1.05 mol x 136.286 g/mol= 143.1 g
the ratio between HCl and H2 is 2 : 1
moles H2 = 1.05
mass H2 =1.05 mol x 2.016 g/mol= 2.12 g
In the second case Zn is the limiting reactant ( 1.21 x 2 = 2.42 moles HCl are needed and we have 2.65 moles HCl)
we get 1.21 moles ZnCl2
mass ZnCl2 = 1.21 x 136.286 g/mol= 164.9 g
and 1.21 moles H2
mass H2 = 1.21 mol x 2.016 g/mol=2.44 g
2008-12-13 8:34 pm
Do you mean Limiting Reactants? Cause Limiting Reagents sounds like nothing ive heard...
收錄日期: 2021-05-01 22:52:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081213122516AA6K5Bg