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]
[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]