Question

It is desired to produce 7.69 grams of nitrogen monoxide by the following reaction. If the percent yield of nitrogen monoxide is 71.9 %, how many grams of oxygen gas would need to be reacted? nitrogen(g) + oxygen( nitrogen monoxide() grams oxygen gas

Answer 1

Answer -

Given,

Mass of Nitrogen monoxide required = 7.69 g

Percentage Yield of Nitrogen Monoxide = 71.9 %

Molar Mass of O2 = 31.9988 g/mol

Molar Mass of NO = 30.0061 g/mol

Mass of Oxygen gas needed = ?

N2 + O2 $\rightarrow$ 2 NO [BALANCED]

Let the mass of O2 required be X g,

Now,

Moles = mass / Molar Mass

So,

Moles of O2 in X grams = X / 31.9988 g/mol

Moles of O2 in X grams = (X/31.9988) mol/g

Now,

Using Stiochiometry, it can be analyzed that for 1 moles of Oxygen gas, 2 moles of Nitrogen Monoxides are produced.

i.e.

Moles of NO = 2 * Moles of O2

So, Moles of NO = 2 * (X/31.9988) mol/g

Moles of NO = 0.0625 * X mol/g

But,

Percentage Yield is 71.9 %.

So,

Actual moles produced = 71.9 % * 0.0625 * X mol/g

Actual moles produced = 0.719 * 0.0625 * X mol/g

Actual moles produced = 0.0449 * X mol/g

Now,

Moles = mass / molar Mass

So,

Mass = Moles * Molar Mass

Mass of NO produced = 0.0449 * X mol/g * 30.0061 g/mol

Mass of NO produced = 1.3472 * X

The required mass is 7.69 g

So,

7.69 g = 1.3472 * X

X = 7.69 g / 1.3472

X = 5.71 g

So, Mass of Oxygen gas required = 5.71 g [ANSWER]