Watch KCV 8.2; read Section 8.3. You can click on the Review link to access the section in your eText.

For the reaction shown, calculate how many moles of [tex]$NO_2$[/tex] form when each amount of reactant completely reacts.

[tex]
2 N_2 O_5(g) \rightarrow 4 NO_2(g) + O_2(g)
[/tex]

Part A

Calculate the moles of [tex]$NO_2$[/tex] formed from [tex]$1.7 \text{ mol } N_2 O_5$[/tex].

Express your answer using two significant figures.

[tex]\boxed{\square \text{ mol } NO_2}[/tex]

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Answer :

To solve this problem, we start with the balanced chemical equation:

[tex]\[ 2 \, \text{N}_2\text{O}_5(g) \rightarrow 4 \, \text{NO}_2(g) + \text{O}_2(g) \][/tex]

This tells us that 2 moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex] produce 4 moles of [tex]\( \text{NO}_2 \)[/tex].

First, determine the mole ratio from the balanced equation. For every 2 moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex], 4 moles of [tex]\( \text{NO}_2 \)[/tex] are produced. Simplifying this ratio:

[tex]\[ \frac{4 \, \text{moles of NO}_2}{2 \, \text{moles of N}_2\text{O}_5} = 2 \, \text{moles of NO}_2 \text{ per mole of N}_2\text{O}_5 \][/tex]

Given that we start with 1.7 moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex], we can use the simplified mole ratio to find the number of moles of [tex]\( \text{NO}_2 \)[/tex]:

[tex]\[ \text{Moles of NO}_2 = 1.7 \, \text{moles of N}_2\text{O}_5 \times 2 \, \text{moles of NO}_2 / \text{moles of N}_2\text{O}_5 \][/tex]

[tex]\[ \text{Moles of NO}_2 = 1.7 \times 2 \][/tex]

[tex]\[ \text{Moles of NO}_2 = 3.4 \][/tex]

Therefore, when 1.7 moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex] completely react, they produce 3.4 moles of [tex]\( \text{NO}_2 \)[/tex].

Expressing the answer with two significant figures we get:

[tex]\[ \boxed{3.4} \][/tex]