# How many grams of dinitrogen pentaoxide can be formed from 15.0 grams of oxygen gas?

## Question:

How many grams of dinitrogen pentaoxide can be formed from 15.0 grams of oxygen gas?

## Balanced Chemical Equations:

The balanced chemical equation for the reaction between substances involves information that can be helpful in computations. The coefficients of the substances provide a relative amount for each substance that completely reacts with another substance in the reaction.

Determine the mass of {eq}\displaystyle N_2O_5 {/eq} that would be produced from the given amount of substance. We consult with the balanced chemical equation,

{eq}\displaystyle 2N_2 + 5O_2 \to 2 N_2O_5 {/eq}

wherein we assume that oxygen gas reacts with an excess of nitrogen gas. We notice that 5 moles of {eq}\displaystyle O_2 {/eq} correspond to 2 moles of {eq}\displaystyle N_2O_5 {/eq}, such that we can relate the number of moles between the substances as

{eq}\displaystyle 2n_{O_2} = 5n_{N_2O_5} {/eq}

Moreover, we express the number of moles as the quotient of the mass, m, and the molar mass, MW, such that

{eq}\displaystyle n = \frac{m}{MW} {/eq}

We use the following values:

• {eq}\displaystyle MW_{O_2} = 32.00\ g/mol {/eq}
• {eq}\displaystyle MW_{N_2O_5} = 108.01 g/mol\ g/mol {/eq}
• {eq}\displaystyle m_{O_2} = 15.0\ g {/eq}

We proceed with finding {eq}\displaystyle m_{N_2O_5} {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle 2n_{O_2} &= 5n_{N_2O_5}\\ 2\frac{m_{O_2}}{MW_{O_2}} &= 5 \frac{m_{N_2O_5}}{MW_{N_2O_5}}\\ \frac{2}{5} \frac{m_{O_2}MW_{N_2O_5}}{MW_{O_2}} &= m_{N_2O_5}\\ \frac{2}{5}\frac{(15.0\ g)(108.01\ g/mol)}{32.00\ g/mol} &= m_{N_2O_5}\\ \boxed{\rm 20.3\ g\ N_2O_5} &\approx m_{N_2O_5} \end{align} {/eq}