Question

If 65.0 g of nitrogen dioxide is reacted with excess water, calculate the theoretical yield

Answer 1

Theoretical Yield of this reaction is 59.34 g. Following is the solution,

Answer 2

Answer: The theoretical yield of the reaction is 14.1 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      ......(1)

Given mass of nitrogen dioxide = 65.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{65.0g}{46g/mol}=1.41mol$

The chemical equation for the reaction of nitrogen dioxide and water follows:

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 1 mole of NO

So, 1.41 moles of nitrogen dioxide will produce = $\frac{1}{3}\times 1.41=0.47mol$ of NO

Now, calculating the mass of NO by using equation 1, we get:

Moles of NO = 0.47 moles

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$0.47mol=\frac{\text{Mass of NO}}{30g/mol}\\\\\text{Mass of NO}=(0.47mol\times 30g/mol)=14.1g$

Hence, the theoretical yield of the reaction is 14.1 grams.