1. Oblicz ile kwasu octowego nalezy spalic aby otrzymac 88 g dwutlęku węgla
2. Oblicz ile kwasu mrówkowego nalezy spalic aby otrzymac 88 g dwutlęku węgla
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zad.1.
CH3COOH + 2O2 -> 2CO2 + 2H2O
M CH3COOH = 2*12 + 4*1 + 2*16 = 60g/mol
2M CO2 = 2*(12+2*16) = 2*44 = 88g/mol
60g CH3COOH ---- 88g CO2
xg ---- 88g CO2
x = 60g CH3COOH
zad.2.
HCOOH + O2 -> CO2 + H2O
M HCOOH = 46g/mol
M CO2 = 44g/mol
46g HCOOH ---- 44g CO2
xg ---- 88g CO2
x = 92g HCOOH
:)
1. CH3COOH + 2O2 -> 2CO2 + 2H2O
M CH3COOH= 60g/mol
2MCO2= 88g/mol
60g CH3COOH-> 88g CO2
xg -> 88g CO2
x = 60g CH3COOH
2. HCOOH + O2 -> CO2 + H2O
M HCOOH = 46g/mol
M CO2 = 44g/mol
46g HCOOH -> 44g CO2
xg -> 88g CO2
x = 92g HCOOH