Methane (CH4) reacts with oxygen gas to form carbon dioxide and water vapour in a complete hydrocarbon combustion reaction.\r\n\r\nAt 21.1 oC and a pressure of 99.650 kPa, 54 L of carbon dioxide are produced. Calculate the mass (in grams) of methane required.

Methane (ch4) reacts with oxygen gas to form carbon dioxide and water vapour in a complete hydrocarbon combustion reaction.rnrnat 21.1 oc and a pressure of 99.650 kpa, 54 l of carbon dioxide are produced. Calculate the mass (in grams) of methane required.

Accepted Answer

Sure, here is the calculation for the mass of methane required in grams:

Step 1: Convert the volume of carbon dioxide from L to moles.

Assuming ideal gas behavior, we can use the ideal gas law:

PV = nRT

where:

P is the pressure in Pa (99.650 kPa = 99650 Pa)

V is the volume in m3 (54 L = 0.054 m3)

n is the number of moles

R is the ideal gas constant (8.314 J/mol*K)

T is the temperature in K (21.1 oC = 294.25 K)

Solving for n, we get:

n = PV/RT

n = (99650 Pa * 0.054 m3) / (8.314 J/mol*K * 294.25 K)

n = 0.215 mol CO2

Step 2: Use the stoichiometry of the reaction to convert moles of carbon dioxide to moles of methane.

The balanced chemical equation for the complete combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we see that 1 mole of CH4 produces 1 mole of CO2.

Therefore, 0.215 moles of CO2 is produced from 0.215 moles of CH4.

Step 3: Convert moles of methane to grams.

The molar mass of CH4 is 16.04 g/mol.

Therefore, 0.215 moles of CH4 is equal to 0.215 moles * 16.04 g/mol = 3.45 g.

Therefore, the mass of methane required is 3.45 g.

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