Question #7b5a1

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Question #7b5a1

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Answer 1 · 2015-11-17T20:03:45

$13 L$ $"13 L"$

Explanation:

The key to this problem is the fact that the pressure and temperature of the gas sample remain constant before and after the chemical reaction takes place.

This means that you can use Avogadro's Law, which states that number of moles and volume have a direct relationship when temperature and pressure are kept constant.

In simple terms, when the number of moles of gas increases, the volume of the gas increases as well. Likewise, when the number of moles of gas decreases, the volume of the gas decreases as well.

![http://www.ltcconline.net/stevenson/2008CHM101Fall/CHM101LectureNotes20081103.htm]( useruploads.socratic.org )

Mathematically, this is written as

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$ $color(blue)(V_1/n_1 = V_2/n_2)" "$ , where

$V_{1}$ $V_1$ , $n_{1}$ $n_1$ - the volume and number of moles at an initial state
$V_{2}$ $V_2$ , $n_{2}$ $n_2$ - the volume and number of moles of gas at a final state

In your case, you know that you initially had $0.11$ $"0.11$ moles of gas. After the reaction is completed, the number of moles of gas increased by $0.58$ $0.58$ moles, to a new value of

$n_{2} = 0.11 + 0.58 = 0.69 moles$ $n_2 = 0.11 + 0.58 = "0.69 moles"$

This means that the new volume of the cylinder will be

$V_{2} = \frac{n_{2}}{n_{1}} \cdot V_{1}$ $V_2 = n_2/n_1 * V_1$

$V_{2} = \frac{0.69 moles}{0.11 moles} \cdot 2.1 L$ $V_2 = (0.69color(red)(cancel(color(black)("moles"))))/(0.11color(red)(cancel(color(black)("moles")))) * "2.1 L"$

$V_{2} = 13.17 L$ $V_2 = "13.17 L"$

Rounded to two sig figs, the number of sig figs you have for the values given, the answer will be

$V_{2} = 13 L$ $V_2 = color(green)("13 L")$

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Question #7b5a1

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