What mass of water will be produce if #11.6*mg# of butane are completely combusted?

1 Answer
Aug 30, 2017

Answer:

Under #20*mg# of water will be produced......

Explanation:

As always, we need a stoichiometrically balanced equation....

#C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)#

And thus #5 *mol# of water result from the combustion of #1*mol# of butane.

#"Moles of butane"=(11.6xx10^-3*g)/(58.12*g*mol^-1)=2.00xx10^-4*mol#

And so the mass of water produced will be.....

#2.00xx10^-4*molxx5xx18.01*g*mol^-1=0.01797*g#

#=0.01797*gxx10^3*mg*g^-1=??*mg#