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missed this? watch kcv: reaction stoichiometry, read section 8.3. you can click on the review link to access the section in your e text. for the reaction shown, calculate how many moles of no₂ form when each amount of reactant completely reacts. 2 n₂o₅ (g) → 4 no₂ (g) + o₂ (g) part d 1.005 × 10⁻³ mol n₂o₅ express your answer using four significant figures.
Step1: Identify mole ratio
From reaction $2\text{N}_2\text{O}_5(g)
ightarrow 4\text{NO}_2(g) + \text{O}_2(g)$, ratio of $\text{NO}_2$ to $\text{N}_2\text{O}_5$ is $4:2 = 2:1$.
Step2: Calculate moles of $\text{NO}_2$
Moles of $\text{NO}_2 = 1.005 \times 10^{-3}\ \text{mol}\ \text{N}_2\text{O}_5 \times 2 = 2.010 \times 10^{-3}\ \text{mol}$
Step3: Round to 4 significant figures
$2.010 \times 10^{-3}$ has 4 significant figures.
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$2.010 \times 10^{-3}$ mol