Questions: If the reaction below absorbs 1,350 kJ of energy, how many grams of N2O will be produced? (N2O=44.0 g / mol) 2 N2(g)+O2(g) -> 2 N2O(g) Delta Hrxn=+163.2 kJ 364 g 8.27 g 5.01 x 10^3 g 728 g

If the reaction below absorbs 1,350 kJ of energy, how many grams of N2O will be produced? (N2O=44.0 g / mol)

2 N2(g)+O2(g) -> 2 N2O(g)
Delta Hrxn=+163.2 kJ

364 g
8.27 g
5.01 x 10^3 g
728 g

Transcript text: If the reaction below absorbs $1,350 \mathrm{~kJ}$ of energy, how many grams of $\mathrm{N}_{2} \mathrm{O}$ will be produced? $\left(\mathrm{N}_{2} \mathrm{O}=44.0 \mathrm{~g} / \mathrm{mol}\right)$ \[ \begin{array}{l} 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \square_{2} \mathrm{~N}_{2} \mathrm{O}(g) \\ \Delta \mathrm{Hrxn}=+163.2 \mathrm{~kJ} \end{array} \] 364 g 8.27 g $5.01 \times 10^{3} \mathrm{~g}$ 728 g

Solution

Solution Steps

Step 1: Understand the Reaction and Given Data

The reaction given is: \[ 2 \mathrm{~N}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}(g) \] The enthalpy change for the reaction, $\Delta \mathrm{Hrxn}$, is $+163.2 \mathrm{~kJ}$. This means that producing 2 moles of $\mathrm{N}_{2} \mathrm{O}$ absorbs $163.2 \mathrm{~kJ}$ of energy.

Step 2: Calculate Moles of $\mathrm{N}_{2} \mathrm{O}$ Produced

We need to find out how many moles of $\mathrm{N}_{2} \mathrm{O}$ are produced when $1,350 \mathrm{~kJ}$ of energy is absorbed. Using the ratio from the enthalpy change: \[ \frac{2 \text{ moles of } \mathrm{N}_{2} \mathrm{O}}{163.2 \mathrm{~kJ}} = \frac{x \text{ moles of } \mathrm{N}_{2} \mathrm{O}}{1,350 \mathrm{~kJ}} \] Solving for $x$: \[ x = \frac{2 \times 1,350}{163.2} = 16.54 \text{ moles of } \mathrm{N}_{2} \mathrm{O} \]

Step 3: Convert Moles to Grams

The molar mass of $\mathrm{N}_{2} \mathrm{O}$ is $44.0 \mathrm{~g/mol}$. To find the mass in grams: \[ \text{Mass} = 16.54 \text{ moles} \times 44.0 \mathrm{~g/mol} = 727.76 \mathrm{~g} \]