What is the minimum mass of oxygen, in grams, needed to burn 16.4 g "C"_4"H"_12"O"?

Give your answer to three significant figures. The unbalanced chemical equation is $\text{C"_4"H"_10"O"+"O"_2->"CO"_2+"H"_2"O}$

Sep 27, 2017

We will use butanol.....

Explanation:

${C}_{4} {H}_{10} O \left(l\right) + 6 {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right)$

Is this balanced with respect to mass and charge? Do not trust my arithmetic!

And so $\text{moles of butanol} = \frac{16.4 \cdot g}{74.12 \cdot g \cdot m o {l}^{-} 1} = 0.221 \cdot m o l$

And so with respect to dioxygen, we need a mass of....

$0.221 \cdot m o l \times 6 \times 32.00 \cdot g \cdot m o {l}^{-} 1 = 42.5 \cdot g$.