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What mass of oxygen is needed for the complete combustion of #8.90 * 10^-3# #g# of methane?

1 Answer

Feb 28, 2016

#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)#

Explanation:

#"Moles of methane"# #=# #(8.90xx10^-3*g)/(16.04*g*mol^-1)# #=# #??# #"moles"#

Given the stoichiometric equation, quite clearly, we need twice this molar quantity of dioxygen gas for complete combustion:

#2xx(8.90xx10^-3*g)/(16.04*g*mol^-1)xx32.00*g*mol^-1" dioxygen gas"#

#=# #??g# #"dioxygen gas"#.

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