Question

Calculate the quantity of O₂ would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 67.1% yield.

2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

Related Concepts

• Mole Ratios and Stoichiometry
• Moles, molar mass and mass
• Percentage yield

Solving the Question

We're given:

• 
  `2 NO_((g)) + O_(2(g))\rightarrow 2 NO_(2 (g))`

• 
  `n_(NO_2)=13.0\hspace{4}mol`
• 
  `\%\hspace{4}yield=67.1\%`

First, let's calculate the theoretical yield of NO2.

⇒
`\%\hspace{4}yield=\frac{actual\hspace{4}yield}{theoretical\hspace{4}yield}`

The yield of NO2 we must produce is 13.0 mol.

⇒
`\%\hspace{4}yield=\frac{13.0}{theoretical\hspace{4}yield}`

We're also given that the percentage yield of this reaction is 67.1%.

⇒
`0.671=\frac{13.0}{theoretical\hspace{4}yield}`

⇒
`theoretical\hspace{4}yield=(13.0)/(0.671)\\\\theoretical\hspace{4}yield=19.3740685544\hspace{4}mol`

First, construct a mole ratio between O2 and NO2.

From the balanced chemical equation, we know that 1 mol of O2 can produce 2 mol of NO2:

⇒
`(n_(O_2))/(1)=(n_(NO_2))/(2)`

We're given that we must theoretically produce 19.3740685544 mol of NO2:

⇒
`(n_(O_2))/(1)=(19.3740685544)/(2)`

⇒
`n_(O_2)=9.6870342772\hspace{4}mol`

Round to significant figures (3):

⇒
`n_(O_2)=9.69\hspace{4}mol`

Answer

`n_(O_2)=9.69\hspace{4}mol`