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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. Are you sure?
Accepted Answer
Sure, here is the calculation for the mass of methane required:
Firstly, convert the volume of carbon dioxide from Litres to cubic centimetres:
V = 54 L * (1000 cm3/1 L) = 54000 cm3
Next, use the ideal gas law to calculate the number of moles of carbon dioxide produced:
PV = nRT
n = PV/RT = (99.650 kPa * 54000 cm3) / (0.0821 Latm/(molK) * 294.25 K)
= 24.46 mol CO2
Since 1 mole of methane produces 1 mole of carbon dioxide, the number of moles of methane required is also 24.46 mol.
Finally, convert the number of moles of methane to mass in grams:
m = n * M = 24.46 mol * 16.04 g/mol
= 392.12 g
Therefore, 392.12 g of methane is required to produce 54 L of carbon dioxide at 21.1 oC and 99.650 kPa.
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