Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
The level of toxins in the fish's cell is equivalent to the level of toxins in the water. Therefore, in order to reduce the toxins further, we should replace the now contaminated water with clean water. After the level of toxins in the fish's cell stops reducing, we replace the water with clean water once again.
When magnesium ion doesn't give any characteristics colour with the flame test as electronic transisitons do not give out visible light.
