To calculate the theoretical yield of carbonic acid ( [tex]\( H_2CO_3 \)[/tex] ) formed when [tex]\( 495 \)[/tex] milliliters of carbon dioxide ([tex]\( CO_2 \)[/tex]) reacts with excess water under the given conditions, we can use the Ideal Gas Law, [tex]\( PV = nRT \)[/tex]. Let's walk through the necessary steps:
1. Convert the Volume to Liters:
Given volume of [tex]\( CO_2 \)[/tex] is [tex]\( 495 \)[/tex] milliliters.
[tex]\[
495 \text{ milliliters} = 495 \times \frac{1 \text{ liter}}{1000 \text{ milliliters}} = 0.495 \text{ liters}
\][/tex]
2. Convert the Temperature to Kelvin:
Given temperature is [tex]\( 25^{\circ} C \)[/tex].
[tex]\[
25^{\circ} C + 273.15 = 298.15 \text{ K}
\][/tex]
3. Use the Ideal Gas Law to Find the Number of Moles ([tex]\( n \)[/tex]) of [tex]\( CO_2 \)[/tex]:
Given pressure [tex]\( P = 101.3 \)[/tex] kilopascals,
Ideal Gas constant [tex]\( R = 8.314 \frac{L kPa}{mol K} \)[/tex],
Temperature [tex]\( T = 298.15 \text{ K} \)[/tex],
Volume [tex]\( V = 0.495 \text{ liters} \)[/tex].
Rearranging the Ideal Gas Law to solve for [tex]\( n \)[/tex]:
[tex]\[
n = \frac{PV}{RT}
\][/tex]
Substitute the given values:
[tex]\[
n = \frac{101.3 \text{ kPa} \times 0.495 \text{ liters}}{8.314 \frac{L kPa}{mol K} \times 298.15 \text{ K}}
\][/tex]
[tex]\[
n \approx 0.02023 \text{ moles}
\][/tex]
4. Calculate the Theoretical Yield of [tex]\( H_2CO_3 \)[/tex]:
According to the balanced chemical equation:
[tex]\[
CO_2 + H_2O \rightarrow H_2CO_3
\][/tex]
1 mole of [tex]\( CO_2 \)[/tex] produces 1 mole of [tex]\( H_2CO_3 \)[/tex]. Thus, moles of [tex]\( H_2CO_3 \)[/tex] formed will be the same as moles of [tex]\( CO_2 \)[/tex], which is [tex]\( 0.02023 \text{ moles} \)[/tex].
5. Calculate the Mass of [tex]\( H_2CO_3 \)[/tex]:
Molar mass of [tex]\( H_2CO_3 \)[/tex] is calculated as follows:
[tex]\[
Molar \, mass \, of \, H_2CO_3 = 2 \times 1 (H) + 12 (C) + 3 \times 16 (O) = 62 \, g/mol
\][/tex]
Mass of [tex]\( H_2CO_3 \)[/tex] formed:
[tex]\[
\text{Mass} = \text{moles} \times \text{molar mass}
\][/tex]
[tex]\[
\text{Mass} = 0.02023 \text{ moles} \times 62 \, \text{g/mol} \approx 1.254 \, \text{g}
\][/tex]
To match the multiple-choice answers:
Option A: [tex]\( 0.889 \, g \)[/tex]
Option B: [tex]\( 1.10 \, g \)[/tex]
Option C: [tex]\( 1.27 \, g \)[/tex]
Option D: [tex]\( 2.029 \, g \)[/tex]
Since [tex]\( 1.254 \)[/tex] grams is closest to [tex]\( 1.27 \)[/tex], the correct answer is:
[tex]\[
\boxed{1.27 \, g}
\][/tex]
