8) what would the formula of di-iodine pentasulfide be?

The formula for di-iodine pentasulfide would be I2S5.

Two (di) molecules of iodine, paired with five (penta) molecules of sulfur.

0.664 g are formed by the reaction.

Explanation:

First of all, we determine the reaction:

Mg + CO₂  →  MgO  + CO

We need to determine the moles of the CO₂ by the Ideal Gases Law.

We convert to L, the volume → 400 mL = 0.4L

T° → 23°C + 273 = 296K

P . V = n . R .T

n = P . V / R .T

n = (1 atm . 0.4L) / (0.082 . 296K) → 0.0165 moles

Moles of Mg → 2.5 g . 1mol / 24.3g = 0.103 moles

As ratio is 1:1, CO₂ is the limiting reactant.

For 1 mol of Mg, we need 1 mol of gas

Then, for 0.103 moles of Mg, we need 0.103 moles of gas, but we only have 0.0165 moles.

If we see the product side, ratio is also 1:1

0.0165 moles of CO₂ must produce 0.0165 moles of MgO.

We convert the moles to mass → 0.0165 mol . 40.3 g /1mol = 0.664 g