Question

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

Answer 1

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`