What is the final volume if 20 L methane reacts completely with 20 L oxygen CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ?
at 100◦C and 2 atmospheres?
1. 20 L
2. 30 L
3. 10 L
4. Cannot be determined from the information given.
5. 15 L
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1 part methane (1x20) reacts with 2 parts oxygen (2x10) to make 1 part carbon dioxide and 2 parts water.
CH4 + 2 O2 → CO2 + 2 H2O
The law of conversion of mass, the mass in will always equal the mass out.
At 100 degrees and 2 atmosphere
20L Oxygen +20 L methan------->40L (CO2's mass+water's mass)
Use the perfect gas regulation: PV = nRT with the aid of fact the strain (P) the gas consistent (R) and the temperature (T) are consistent, the quantity might desire to be course proportional to the style of moles of gas. this type, you could cope with the style of liters of a gas purely like the style of moles of a gas. So, if 20L of CH4 reacted, two times that plenty, 40L, of O2 could be mandatory, yet for the reason that there's no longer that plenty O2, O2 is the proscribing reactant. for this reason, for each 2 moles of O2 reacted, a million mole of CO2 consequences, 20L of O2 might consequence in 10L of CO2. this might mean that there's no O2 left. yet, for the reason that 20L of O2 reacted, 10L of CH4 might desire to have reacted, leaving 10L of CH4 last. So, the only gases that are left are 10L of CH4 and 10L of CO2, so the great quantity is 10L + 10L = 20L =]