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 dа ta: 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 dа ta: 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