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

Aug 30, 2017

Under $20 \cdot m g$ of water will be produced......

#### Explanation:

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

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right)$

And thus $5 \cdot m o l$ of water result from the combustion of $1 \cdot m o l$ of butane.

$\text{Moles of butane} = \frac{11.6 \times {10}^{-} 3 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1} = 2.00 \times {10}^{-} 4 \cdot m o l$

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

$2.00 \times {10}^{-} 4 \cdot m o l \times 5 \times 18.01 \cdot g \cdot m o {l}^{-} 1 = 0.01797 \cdot g$

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