Question

Given the following reaction: ___ NO (g) + ___O2 (g)→___NO2 (g)How many liters of gaseous oxygen are needed to produce 4.50 L of gaseous nitrogen dioxide, if both gases are being measured at STP?

Answer 1

2.24 L

Step-by-step explanation

Given:

volume of nitrogen dioxide produced = 4.50 L

The given unbalanced equation is;

___ NO (g) + ___O2 (g) → ___NO2 (g)

What to find:

The liters of gaseous oxygen needed to produce 4.50 L of gaseous nitrogen dioxide at STP.

Step-by-step solution:

The first step is to write a balanced equation for the reaction.

2 NO (g) + 1 O2 (g) → 2 NO2 (g)

The next step is to convert 4.5 L NO2 to moles.

Conversion factor: 1 mole of any gas at STP occupies 22.4L.

Therefore, the number of moles of NO2 in 4.5 L volume will be

`(4.5L)/(22.4L)*1mol=0.20\text{ }mol\text{ }NO_2`

The next step is to calculate the mole of O2 needed to produce 0.20 mol NO2.

From the balanced equation 1 mol O2 produce 2 mol NO2

So, the x mol O2 will be needed to prduce 0.20 mol NO2

`\frac{0.20mol\text{ }NO_2}{2mol\text{ }NO_2}*1mol\text{ }O_2=0.1mol\text{ }O_2`

The final step is to convert 0.1 mol O2 to liters using the conversion factor in step 2.

The liters of gaseous oxygen needed is

`(0.1mol)/(1mol)*22.4L=2.24\text{ }L`

The liters of gaseous oxygen needed to produce 4.50 L of gaseous nitrogen dioxide at STP = 2.24 L