Ile gramów fenolu przearagowało z sodem jesli wytworzyło się 5,6 dm³ H₂ w warunkach normalnych?
1mol x
------ = -------
22,4dm³ 5,6dm³
x = 0,25mola
2C₆H₅OH + 2Na --> 2C₆H₅ONa + H₂
188g fenolu x
----------- =-----------
1 mol wodoru 0,25mola wodoru
x = 47g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
M(C₆H₅OH)=6*12g/mol+6*1g/mol+16g/mol=94g/mol
2C₆H₅OH+2Na-->2C₆H₅ONa+H₂
2*94g---------------------22,4dm³
xg------------------------5,6dm³
xg=188g*5,6dm³/22,4dm³
xg=47g
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
1mol x
------ = -------
22,4dm³ 5,6dm³
x = 0,25mola
2C₆H₅OH + 2Na --> 2C₆H₅ONa + H₂
188g fenolu x
----------- =-----------
1 mol wodoru 0,25mola wodoru
x = 47g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
M(C₆H₅OH)=6*12g/mol+6*1g/mol+16g/mol=94g/mol
2C₆H₅OH+2Na-->2C₆H₅ONa+H₂
2*94g---------------------22,4dm³
xg------------------------5,6dm³
xg=188g*5,6dm³/22,4dm³
xg=47g