Ile tlenu należy zmieszać z 6g wodoru w celu otrzymania .??
m H₂ = 6g
M H₂ = 2g/mol
n = m/M
n = 6g/2g/mol = 3mole
2H₂ + O₂ --> 2H₂O
2mole H₂ 1mol O₂
----------- = ------------
3mole H₂ x
x = 1,5mola O₂
1 mol 22,4dm³
------ = --------------
1,5mol x
x = 33,6dm³
Pozdrawiam Marco12 ;)
2H2 + O2 --> 2H2O
2*22,4dm^3 --- 22,4 dm^3
3*22,4dm^3 ---- x
x = (3*22,4*22,4)/ (2*22,4) = (67,2*22,4)/44,8 = 1505,28/44,8 = 33,6 dm^3
1 mol = 22,4 dm^3
W razie pytań zapraszam na PW :)
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m H₂ = 6g
M H₂ = 2g/mol
n = m/M
n = 6g/2g/mol = 3mole
2H₂ + O₂ --> 2H₂O
2mole H₂ 1mol O₂
----------- = ------------
3mole H₂ x
x = 1,5mola O₂
1 mol 22,4dm³
------ = --------------
1,5mol x
x = 33,6dm³
Pozdrawiam Marco12 ;)
2H2 + O2 --> 2H2O
2*22,4dm^3 --- 22,4 dm^3
3*22,4dm^3 ---- x
x = (3*22,4*22,4)/ (2*22,4) = (67,2*22,4)/44,8 = 1505,28/44,8 = 33,6 dm^3
1 mol = 22,4 dm^3
W razie pytań zapraszam na PW :)