Question

missed this? watch kcv: reaction stoichiometry, me: stoichiometry, click on the review link to access the section in your etext. balance the equation and calculate how many moles of o₂ form when each quantity of reactant completely reacts. n₂o₅(g)→no₂(g)+o₂(g) part b 1.7 mol n₂o₅ express your answer in moles to two significant figures. view available hint(s) part c 3.8 mol n₂o₅ express your answer in moles to two significant figures.

Explanation:

Step1: Balance the chemical equation

The un - balanced equation is $N_2O_5(g)
ightarrow NO_2(g)+O_2(g)$. By balancing the nitrogen and oxygen atoms, we get $2N_2O_5(g)
ightarrow 4NO_2(g)+O_2(g)$.

Step2: Use stoichiometry for part B

The balanced equation shows that the mole ratio of $N_2O_5$ to $O_2$ is 2:1. Given $n_{N_2O_5}=1.7$ mol. Using the ratio $\frac{n_{O_2}}{n_{N_2O_5}}=\frac{1}{2}$, we have $n_{O_2}=\frac{1}{2}\times n_{N_2O_5}$. Substituting the value of $n_{N_2O_5}$, we get $n_{O_2}=\frac{1}{2}\times1.7$ mol $ = 0.85$ mol.

Step3: Use stoichiometry for part C

We know from the balanced equation that for every 2 moles of $N_2O_5$ that react, 1 mole of $O_2$ is produced. Given we start with 3.8 mol of $N_2O_5$. Using the mole - ratio $\frac{n_{O_2}}{n_{N_2O_5}}=\frac{1}{2}$, we calculate $n_{O_2}=\frac{1}{2}\times3.8$ mol $=1.9$ mol.

Answer:

Part B: $n = 0.85$ mol
Part C: $n = 1.9$ mol