Answer: 25.1195 L
Explanation:
<u>Given:</u>
There are 0.5860 mol of Iron powder (Fe)
The volume of oxygen gas at STP is 31.5 L
The volume filled by a gas at STP is approximately 22.71 L
Therefore, the moles of oxygen gas at STP = Volume of oxygen gas at STP/ 22.71
= 35.1/22.71
= 1.5456 mol
The balanced chemical equation given for this reaction is:
4Fe(s) + 3O2 (g) ----> 2Fe2O3 (s)
Dividing the above equation by 2, yields:
2Fe(s) + 3/2O2(g) -------> Fe2O3 (s)
From the equation we see:
2 moles of Fe reacts with 3/2 mole of O2 to give one mole of Fe2O3
Thus, let x moles of O2 react with 0.5860 moles of Fe
0.5860/2=x/3/2
0.5860*3=4*x
x = 1.758/4
x =0.4395 mol
So the remaining moles of oxygen gas = 1.5456 - 0.4395
= 1.1061 mol
Therefore, the amount of oxygen gas, in litres, remaining is = 1.1061*22.71
= 25.1195 L
