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# Using 50g I?O? in Respirators: CO Removal Capacity & Remaining I? Mass

**Chapter 6, Problem 6.23**

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**QUESTION:**

Diiodine pentoxide is used in respirators to remove carbon monoxide from air:

\(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(\mathrm{~g})\)

(a) What mass of carbon monoxide could be removed from air by a respirator that contains 50.0 g of diiodine pentoxide?

(b) What mass of \(\mathrm{I}_{2}\) would remain in the respirator?

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### Questions & Answers

**QUESTION:**

Diiodine pentoxide is used in respirators to remove carbon monoxide from air:

\(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(\mathrm{~g})\)

(a) What mass of carbon monoxide could be removed from air by a respirator that contains 50.0 g of diiodine pentoxide?

(b) What mass of \(\mathrm{I}_{2}\) would remain in the respirator?

**ANSWER:**

Step 1 of 3

(a) The given reaction is as follows;

\(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \rightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\)

\(\begin{array}{l}\text{Molar mass of } \mathrm{I}_{2} \mathrm{O}_{5}=333.81 \mathrm{~g} / \mathrm{mol}\\ \text{Given mass } =50.0 \mathrm{~g}\end{array}\)

Let’s calculate the number of moles of diiodine:

\(\text { Number of moles of } \mathrm{I}_{2} \mathrm{O}_{5}=\frac{\text { Given mass }}{\text { Molar mass }}\)

From the reaction, \(1 \mathrm{~mol} \text { of } \mathrm{I}_{2} \mathrm{O}_{5} \text { reacts with } \rightarrow 5 \mathrm{~mol} \mathrm{CO}\)

The conversion factor will be:

\(\frac{5 \mathrm{~mol} \mathrm{CO}}{1 \mathrm{~mol}_{2} \mathrm{O}_{5}}\)