Answer:
643g of methane will there be in the room
Step-by-step explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -Assuming STP-. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
Volume Methane:
3600s * (0.25L / s) = 900L Methane
Moles methane:
PV = nRT; PV / RT = n
Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
Mass methane:
40.18 moles * (16g/mol) =
643g of methane will there be in the room