Asked by Dani
The combustion of methane can be described by the equation:
CH4 (g) + 2 O2 (g) -> CO2 (g) +2H20 (g)
If 22.0 mL of CO2 (g) were collected, what volume of H20 (g) also could be collected under the same conditions?
CH4 (g) + 2 O2 (g) -> CO2 (g) +2H20 (g)
If 22.0 mL of CO2 (g) were collected, what volume of H20 (g) also could be collected under the same conditions?
Answers
Answered by
DrBob222
22.0 mL CO2 = how many moles. That is 22.0 mL x (1 mol/22,400 ml) = ? mol CO2. Now convert that to mole H2O.
?mole CO2 x (2 moles H2O/1 mole CO2) = ? mole CO2 x 2/1 and that x 22,400 mL/mol will convert to mL H2O.
There is a shortcut you can use when working will all gases. You can work in L and never convert to moles (notice with my work above that I divided by 22,400 mL/mol to convert moles CO2 to moles H2O, then multiplied by 22,400 to convert moles H2O to mL H2O.
The shortcut, using just mL is
22.0 mL CO2 x (2L H2O/1 L CO2) = 44.0 mL H2O(g)
?mole CO2 x (2 moles H2O/1 mole CO2) = ? mole CO2 x 2/1 and that x 22,400 mL/mol will convert to mL H2O.
There is a shortcut you can use when working will all gases. You can work in L and never convert to moles (notice with my work above that I divided by 22,400 mL/mol to convert moles CO2 to moles H2O, then multiplied by 22,400 to convert moles H2O to mL H2O.
The shortcut, using just mL is
22.0 mL CO2 x (2L H2O/1 L CO2) = 44.0 mL H2O(g)
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