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When 1.0 mole of Cu2FeS3(s) and 1.0 mole of O2(g) are mixed and allowed to react according to the equation 2Cu2FeS3(s) + 7O2(g) → 6Cu(s) + 2Fe2O3(s) + 6SO2(g) - VCE - SSCE Chemistry - Question 4 - 2010 - Paper 1
Question 4
When 1.0 mole of Cu2FeS3(s) and 1.0 mole of O2(g) are mixed and allowed to react according to the equation 2Cu2FeS3(s) + 7O2(g) → 6Cu(s) + 2Fe2O3(s) + 6SO2(g)
Worked Solution & Example Answer:When 1.0 mole of Cu2FeS3(s) and 1.0 mole of O2(g) are mixed and allowed to react according to the equation 2Cu2FeS3(s) + 7O2(g) → 6Cu(s) + 2Fe2O3(s) + 6SO2(g) - VCE - SSCE Chemistry - Question 4 - 2010 - Paper 1
Step 1
no reagent is in excess.
Answer
To determine if any reagents are in excess, we start by considering the stoichiometry of the reaction. According to the balanced equation, 2 moles of Cu2FeS3 react with 7 moles of O2. Thus, for every mole of Cu2FeS3, we need 3.5 moles of O2.
Given we have 1.0 mole of Cu2FeS3, the amount of O2 needed is:
Since we have 1.0 mole of O2, we can see that O2 is insufficient, meaning that O2 is not in excess.
Step 2
Step 3
5/7 mole of Cu2FeS3 is in excess.
Answer
In this case, we need to consider how many moles of Cu2FeS3 would be consumed based on the available O2. Since we only have 1 mole of O2, the moles of Cu2FeS3 that can react with it is:
1.0 ext{ mole O2} imes rac{2 ext{ moles Cu2FeS3}}{7 ext{ moles O2}} = rac{2}{7} ext{ moles Cu2FeS3}
Thus, out of the initial 1.0 mole of Cu2FeS3, the remaining unreacted amount is:
1.0 - rac{2}{7} = rac{5}{7} ext{ moles Cu2FeS3}
This means 5/7 of a mole of Cu2FeS3 is in excess, making this statement valid.
Step 4