The concentration of methane in an interstitial water samplewas found to be 150

Question

The concentration of methane in an interstitial water samplewas found to be 150 mL/L at STP (standard temperature/pressure).Assuming that the methane was produced by the fermentation oforganic matter, [CH2O], what weight of organic matterwas required to produce the methane in a liter of the interstitialwater? 2 CH2O ---> CH4 (g) +CO2 (g) The concentration of methane in an interstitial water samplewas found to be 150 mL/L at STP (standard temperature/pressure).Assuming that the methane was produced by the fermentation oforganic matter, [CH2O], what weight of organic matterwas required to produce the methane in a liter of the interstitialwater? 2 CH2O ---> CH4 (g) +CO2 (g)

Explanation / Answer

                                                         2CH2O ---> CH4 (g) +CO2 (g) Concentration of methane = 150 mL / L 22.4 L of any gas at STP are equal to 1 mole 150 mL of methane at STP = 1 mole * 0.15 L / 22.4L                                           = 0.0067 moles From the equation, we can see 2 moles of organic matter give 1mole of methane. So 0.0067 moles of methane can be produced by 0.0067 * 2 =0.0134 moles of CH2O So mass of organic matter = Moles * molar mass                                         =0.0134 * 30.02 g/mol                                        = 0.402 g So 0.402 g of organic matter will produce 150 mL of methaneper liter of water.                                                          2CH2O ---> CH4 (g) +CO2 (g) Concentration of methane = 150 mL / L 22.4 L of any gas at STP are equal to 1 mole 150 mL of methane at STP = 1 mole * 0.15 L / 22.4L                                           = 0.0067 moles From the equation, we can see 2 moles of organic matter give 1mole of methane. So 0.0067 moles of methane can be produced by 0.0067 * 2 =0.0134 moles of CH2O So mass of organic matter = Moles * molar mass                                         =0.0134 * 30.02 g/mol                                        = 0.402 g So 0.402 g of organic matter will produce 150 mL of methaneper liter of water.

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