91Ó°ÊÓ

The chemical equation for the process is,

$\mathbf{2}\mathbf{LiOH}\left(s\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{→}{\mathbf{Li}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

From the above reaction, it is concluded that 2 moles of LiOH reacts with 1 mol of CO₂.

Molar mass of 1 mol of LiOH = 23.95 g

Molar mass of CO₂= 44.01 g

So, 47.9 g of LiOH reacts with 44.01 g of CO₂.

Mass of given LiOH = 3.50 kg = 3.50x10³g

Now, the mass of CO₂ is,

$$\begin{gathered} \mathbf{=}\frac{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{×}\left(3.50×{10}^{3}g\right)}{\mathbf{47}\mathbf{.}\mathbf{9}\mathbf{g}} \\ \mathbf{=}\mathbf{3}\mathbf{.}\mathbf{22}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{g} \end{gathered}$$

Thus, the mass of CO₂ that can be removed by reaction with 3.50 kg of lithium hydroxide is 3.22x10³ g.

The chemical equation for the process is,

$\mathbf{2}\mathbf{LiOH}\left(s\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{→}{\mathbf{Li}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

From the above reaction, it is concluded that 2 moles of LiOH reacts with 1 mol of CO₂.

So, 47.9 g of LiOH reacts with 44.01 g of CO₂.

Mass of given LiOH = 1.0 g

Now, the mass of CO₂ is,

$$\begin{gathered} \mathbf{=}\frac{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{×}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{g}}{\mathbf{47}\mathbf{.}\mathbf{9}\mathbf{g}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{919}\mathbf{g} \end{gathered}$$

Hence, the mass of CO₂ can be removed by the reaction with 1.0 g of Lithium hydroxide is 0.919 g.

The chemical equation for the process is,

$\mathbf{Mg}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\left(s\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{→}{\mathbf{MgCO}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

From the above reaction, it is concluded that 1 moles of Mg(OH)₂ reacts with 1 mol of CO₂.

So, 58.33 g of Mg(OH)₂ reacts with 44.01 g of CO₂.

Mass of givenMg(OH)₂ = 1.0 g

Now, the mass of CO₂ is,

$$\begin{gathered} \mathbf{=}\frac{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{×}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{g}}{\mathbf{58}\mathbf{.}\mathbf{33}\mathbf{g}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{755}\mathbf{g} \end{gathered}$$

Hence, the mass of CO₂ can be removed by the reaction with 1.0 g of Magnesium hydroxide is 0.755 g.

The chemical equation for the process is,

$\mathbf{2}\mathbf{AI}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{3}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{→}{\mathbf{AI}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

From the above reaction, it is concluded that 2 moles of Al(OH)₃ reacts with 3mol of CO₂.

So, 156.01 g of Al(OH)₃ reacts with 132.03 g of CO₂.

Mass of given Al(OH)₃ = 1.0 g

Now, the mass of CO₂ is,

$$\begin{gathered} \mathbf{=}\frac{\mathbf{132}\mathbf{.}\mathbf{03}\mathbf{g}\mathbf{×}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{g}}{\mathbf{156}\mathbf{.}\mathbf{01}\mathbf{g}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{846}\mathbf{g} \end{gathered}$$

Hence, the mass of CO₂ can be removed by the reaction with 1.0 g of Aluminum hydroxide is 0.846 g.