Answer:
105.33Kg of O2.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
C55H104O6 + 78O2 —> 55CO2 + 52H2O
Step 2:
Determination of the masses of C55H104O6 and O2 that reacted from the balanced equation.
This is illustrated below:
Molar mass of C55H104O6 = (12x55) + (104x1) + (16x6) = 860g/mol
Mass of C55H104O6 from the balanced equation = 1 x 860 = 860g.
Divide the mass of C55H104O6 by 1000 to express in kg i.e
Mass of C55H104O6 = 860/1000= 0.86Kg
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 78 x 32 = 2496g.
Divide the mass of O2 by 1000 to express in kg i.e
Mass of O2 = 2496/1000 = 2.496kg.
From the balanced equation above,
0.86kg of C55H104O6 reacted with 2.496Kg of O2.
Step 3:
Determination of the mass of O2 in kg needed to react with 36.29 kg of triglycerides (C55H104O6).
This can be obtained as follow:
From the balanced equation above,
0.86kg of C55H104O6 reacted with 2.496Kg of O2.
Therefore, 36.29 kg of C55H104O6 will react with = (36.29 x 2.496)/0.86 = 105.33Kg of O2.
Therefore, 105.33Kg of O2 is needed for the reaction.
