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5.00 g of [tex]$C_6H_{12}O_6$[/tex] reacts completely with molten [tex]$KClO_3$[/tex].

Use your answer to (d)(1) to calculate the total volume of gas released at 630 K and [tex][tex]$1.00 \times 10^5$[/tex] Pa[/tex] in this reaction.

(If you were unable to answer (d)(1), use [tex]$0.0463$ m^3[/tex] in this question. This is not the correct answer to (d)(1).)

Total volume of gas released = [tex]\qquad m^3[/tex]

Sagot :

GinnyAnsweridn
To solve this problem, we'll first need to confirm that we understand the reaction and how the gases are generated. However, since we have already determined an important intermediate result in a previous step, we'll utilize that to find the volume of gas released.

Here, we are given:

- The volume of gas released is [tex]\( 0.0463 \)[/tex] cubic meters.
- Temperature [tex]\( T = 630 \)[/tex] Kelvin.
- Pressure [tex]\( P = 1.00 \times 10^5 \)[/tex] Pascals.

Since we're using our derived result directly for the volume of gas released in this context, the steps can be summarized as follows:

1. Identify the volume of gas released from the previous calculations.
2. State this volume as the total volume of gas released.

Thus, the total volume of gas released at 630 K and [tex]\(1.00 \times 10^5\)[/tex] Pa in this reaction is:

[tex]\[ \text{Total volume of gas released} = 0.0463 \, \text{m}^3 \][/tex]

So, the final total volume of gas released is [tex]\( 0.0463 \, \text{m}^3 \)[/tex].
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