Answer:
a. 324 mL is the volume of CO₂ measured
b. 0.223 moles of carbonate
Explanation:
a. We determine the moles of used O₂ by the Ideal Gases Law
STP are 1 atm of pressure and 273.15K of T°
We convert the volume from mL to L → 0.465 mL . 1L/ 1000 mL = 0.465 L
Now, we replace data: 1 atm . 0465L = n . 0.082 . 273.15K
1 atm . 0465L / 0.082 . 273.15K = n → 0.0207 moles
The balanced combustion is: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
Ratio is 5:3. 5 moles of oxygen are needed to produce 3 moles of CO₂
Then 0.0207 moles must produce (0.0207 . 3) / 5 = 0.0124 moles of CO₂
Let's apply again the Ideal Gases Law. Firstly we convert:
37°C + 273.15K = 310.15K
98.59kPa . 1atm / 101.3 kPa = 0.973 atm
0.973atm . V = 0.0124 mol . 0.082 . 310.15K
V = (0.0124 mol . 0.082 . 310.15K) / 0.973atm = 0.324 L → 324 mL
b. The reaction is: CaCO₃(s) → CaO(s) + CO₂(g)
We used the Ideal Gases Law to determine the moles of produced CO₂
P . V = n . R . T → P .V / R . T = n
We replace data: 1 atm . 5L / 0.082 . 273.15K = 0.223 moles
As ratio is 1:1, 0.223 moles of CO₂ were produced by the decomposition of 0.223 moles of carbonate
