Question

how many grams of oxygen gas (o₂) are required to make 8.2 moles of water (h₂o)?
2 c₂h₂ + 5 o₂ → 2 h₂o + 4 co₂
grams of o₂ =

Explanation:

Step1: Determine mole ratio of \(O_2\) to \(H_2O\)

From the reaction \(2 C_2H_2 + 5 O_2
ightarrow 2 H_2O + 4 CO_2\), the mole ratio of \(O_2\) to \(H_2O\) is \(\frac{5\space moles\space O_2}{2\space moles\space H_2O}\).

Step2: Calculate moles of \(O_2\)

Given moles of \(H_2O = 8.2\space moles\), moles of \(O_2 = 8.2\space moles\space H_2O\times\frac{5\space moles\space O_2}{2\space moles\space H_2O}\)

$$= \frac{8.2\times5}{2}\space moles\space O_2 = 20.5\space moles\space O_2$$

Step3: Convert moles of \(O_2\) to grams

Molar mass of \(O_2\) is \(31.998\space g/mol\). Mass of \(O_2 = 20.5\space moles\times31.998\space g/mol\)

$$= 20.5\times31.998 = 655.959\space g\approx656.0\space g$$

(or as per the given \(31.998\) and calculation, it matches the provided \(655.96\space g\))

Answer:

The mass of oxygen gas required is \(\boldsymbol{655.96\space g}\) (or approximately \(656\space g\))