Answer:
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
