Part A According to avogadro\'s law 1 mole of gas at STP occupy volume = 22.414

Question

Part A

According to avogadro's law 1 mole of gas at STP occupy volume = 22.414 L then in 1.00 X 106 L volume contain mole = 1.00 X 106 / 22.414 = 44615 mole

Tank contain 44615 mole of methane gas

Part B

molar mass of methane = 16.04 gm/mole then 44615 mole of methane = 44615 X 16.04 = 715625 gm of methane

tank contain 715625 gm of methane

Part C

According to avogadro's law 1 mole of gas at STP occupy volume = 22.414 L then in 1.00 X 106 L volume contain mole = 1.00 X 106 / 22.414 = 44615 mole

44615 mole of CO2 in tank

molar mass of CO2 = 44.01 gm/mole then 44615 mole of CO2 = 44615 X 44.01 = 1963506 gm of CO2

tank contain 1963506 gm of Carbon dioxide

Explanation / Answer

Pt. A: How many moles of methane gas, CH4, are in a 1.00×106 L storage tank at STP? Pt. B: How many grams of methane is this? Pt. C: How many grams of carbon dioxide gas could the same tank hold?

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.