Propane burns according to the following equation:
C3H8(g) + O2(g) → CO2(g) + H2O(g)
If 465 mL of oxygen at STP is used in the reaction, what volume of CO2, measured at 37.0oC and 98.59 kPa.
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When calcium carbonate is heated, it produces calcium oxide and carbon dioxide, according to the following reaction:
CaCO3(s) → CaO(s) + CO2(g)
How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP?

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