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

The balanced equation indicates that for 2 moles of Cu2FeS3, 7 moles of O2 are required. For 1 mole of Cu2FeS3, 3.5 moles of O2 are needed, and we only have 1 mole available. Thus:

Excess O2 = Available O2 - Required O2 = 1 - 3.5 = -2.5

Since we have a negative value, this means O2 is not in excess, so this option is incorrect.

We established that 1 mole of Cu2FeS3 requires 3.5 moles of O2 and we have 1 mole of O2 present. To find the excess of Cu2FeS3 after the reaction, we can set up the following:

• We need 3.5 moles of O2 for 1 mole of Cu2FeS3.
• We can only consume 1 mole of O2 available; hence:

Moles of Cu2FeS3 consumed: rac{1}{3.5} = \frac{2}{7}

Thus, we have: $1 - \frac{2}{7} = \frac{5}{7}$

So, rac{5}{7} moles of Cu2FeS3 remain in excess, confirming this option is correct.

We already calculated the moles of Cu2FeS3 consumed and established that rac{2}{7} moles were used up from the 1 mole originally present. The moles remaining are:

$1 - \frac{2}{7} = \frac{5}{7}$

As such, stating that rac{2}{7} moles of Cu2FeS3 are in excess is incorrect.