Answer:
3.38 moles of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4CH2OH + 5O2 → 4CO2 + 6H2O
From the balanced equation above,
4 moles of CH2OH required 5 moles of O2 for complete combustion.
Finally, we shall determine the number of mole of O2 needed to react with 2.7 moles of CH2OH. This can be obtained as follow:
From the balanced equation above,
4 moles of CH2OH required 5 moles of O2 for complete combustion.
Therefore, 2.7 moles of CH2OH will require = (2.7 × 5)/4 = 3.38 moles of O2 for complete combustion.
Thus, 3.38 moles of O2 is required.
Answer:
92.93 g
Explanation:
Number of half lives that have elapsed in eight days =8/14.3 = 0.559
Fraction of the radioactive nuclide that remains after 0.559 half lives is given by
N/No=(1/2)^0.559
Where N= mass of radioactive nuclides remaining after a time t
No= mass of radioactive nuclides originally present
N/No=(1/2)^0.559= 0.679
Mass of nuclides present eight days before= 63.1g/0.679
Mass of nuclides present eight days before=92.93 g
