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

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Answer 1 · 2017-08-30T04:25:07

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

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.....

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

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

What mass of water will be produce if 11.6*mg11.6*mg of butane are completely combusted?