Answer:
Yard . I hope this helped:))
127.88 grams of ethanol were present at the beginning of the reaction
Explanation:
Firstly, let's make the combustion reaction:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
By 1 mol of ethanol, you can make 3 mole of water.
Mole of water = Water mass / Molar mass
150g / 18g/m = 8.3 mole
3 mole of water came from 1 mol of ethanol
8.3 mole came from (8.3 .1)/3 = 2.78 mole of ethanol
Molar mass ethanol = 46 g/m
Mole . molar mass = mass
2.78 m . 46g/m = 127.88 g
First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2
Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2
Beaker 3 shows a clear chemical reaction.
