# What mass of oxygen is needed to react with 4.5g of ethane C_2H_8?

Oct 9, 2016

We need approx. $17 \cdot g$ of dioxygen.

#### Explanation:

A balanced chemical equation is required:

${H}_{3} C - C {H}_{3} + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(g\right)$

And thus each equiv ethane requires $\frac{7}{2}$ equiv of dioxygen.

$\text{Moles of ethane}$ $=$ $\frac{4.5 \cdot g}{30.07 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.15 \cdot m o l$

And thus we need:

$\frac{7}{2} \times 0.15 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1 \text{ dioxygen gas}$ $=$ ??g