iron (IV)sulfide reacts with oxygen gas to form iron III oxide and sulfur dioxid
ID: 1026472 • Letter: I
Question
iron (IV)sulfide reacts with oxygen gas to form iron III oxide and sulfur dioxide gas. ( Note that this reaction is not occurring in water) A) write a balance chemical equation for this reaction including phase labels.B)determine how many moles of sulfur dioxide are formed from the reaction of 96.7 g of iron (IV) sulfide and 55.0 L of oxygen gas ( at 398K and 1.20 atm) iron (IV)sulfide reacts with oxygen gas to form iron III oxide and sulfur dioxide gas. ( Note that this reaction is not occurring in water) A) write a balance chemical equation for this reaction including phase labels.
B)determine how many moles of sulfur dioxide are formed from the reaction of 96.7 g of iron (IV) sulfide and 55.0 L of oxygen gas ( at 398K and 1.20 atm) iron (IV)sulfide reacts with oxygen gas to form iron III oxide and sulfur dioxide gas. ( Note that this reaction is not occurring in water) A) write a balance chemical equation for this reaction including phase labels.
B)determine how many moles of sulfur dioxide are formed from the reaction of 96.7 g of iron (IV) sulfide and 55.0 L of oxygen gas ( at 398K and 1.20 atm)
Explanation / Answer
A) The reaction is as follows:
4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 .
B) Firstly, convert 55L of Oxygen to the estimated volume at STP.
Therefore,
P1V1/T1 = P2V2/T2.
1.2*55/398 = 1*V/273
V= 45.27L.
We know that 480g of FeS2 reacts with 11*22.4L of Oxygen to give 8 moles of Sulphur Dioxide.
11*22.4L Oxygen reacts with 88g of FeS2 ONLY. Thus, Oxygen is the limiting reagent in this question. So, earlier 11*22.4L oxygen gave 8 moles of SO2. Now, 45.27L oxygen will give = 8*45.27/(11*22.4) moles of SO2 = 1.46 moles of SO2.
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