QuestionApril 17, 2026

8.1 A chemist performs the following reaction to make sodium nitrate: 2NaNO_(2)(s)+O_(2)(g)arrow 2NaNO_(3)(s) How many moles of oxygen gas must the chemist use to make 45.0 moles of NaNO_(3) How many moles of NaNO_(2) are necessary?

8.1 A chemist performs the following reaction to make sodium nitrate: 2NaNO_(2)(s)+O_(2)(g)arrow 2NaNO_(3)(s) How many moles of oxygen gas must the chemist use to make 45.0 moles of NaNO_(3) How many moles of NaNO_(2) are necessary?
8.1
A chemist performs the following reaction to make sodium nitrate:
2NaNO_(2)(s)+O_(2)(g)arrow 2NaNO_(3)(s)
How many moles of oxygen gas must the chemist use to make 45.0 moles of NaNO_(3)
How many moles of NaNO_(2) are necessary?

Solution
4.5(307 votes)

Answer

O_2: 22.5 \ \text{mol} ### NaNO_2: 45.0 \ \text{mol} Explanation 1. Determine mole ratio from reaction From 2NaNO_2 + O_2 \rightarrow 2NaNO_3, ratio NaNO_3 : O_2 = 2 : 1 and NaNO_3 : NaNO_2 = 2 : 2. 2. Calculate moles of O_2 45.0 \ \text{mol} \ NaNO_3 \times \frac{1\ \text{mol} \ O_2}{2\ \text{mol} \ NaNO_3} = 22.5 \ \text{mol} \ O_2. 3. Calculate moles of NaNO_2 45.0 \ \text{mol} \ NaNO_3 \times \frac{2\ \text{mol} \ NaNO_2}{2\ \text{mol} \ NaNO_3} = 45.0 \ \text{mol} \ NaNO_2.

Explanation

1. Determine mole ratio from reaction<br /> From $2NaNO_2 + O_2 \rightarrow 2NaNO_3$, ratio $NaNO_3 : O_2 = 2 : 1$ and $NaNO_3 : NaNO_2 = 2 : 2$.<br />2. Calculate moles of $O_2$<br /> $45.0 \ \text{mol} \ NaNO_3 \times \frac{1\ \text{mol} \ O_2}{2\ \text{mol} \ NaNO_3} = 22.5 \ \text{mol} \ O_2$.<br />3. Calculate moles of $NaNO_2$<br /> $45.0 \ \text{mol} \ NaNO_3 \times \frac{2\ \text{mol} \ NaNO_2}{2\ \text{mol} \ NaNO_3} = 45.0 \ \text{mol} \ NaNO_2$.
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