Q134CP - Chemistry: Molecular Nature Of Matter And Change

Question

Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react with each other to produce CO₂, which causes dough or batter to rise:

$\mathbf{Ca}{\left(H_2{\mathrm{PO}}_4\right)}_{\mathbf{2}}\left(s\right)\mathbf{+}{\mathbf{NaHCO}}_{\mathbf{3}}\left(s\right)\mathbf{\to }{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}{\mathbf{CaHPO}}_{\mathbf{4}}\left(s\right)\mathbf{+}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}\left(s\right)$

 [unbalanced]

If the baking powder contains 31% NaHCO₃and 35% Ca(H₂PO₄)₂ by mass:

(a) How many moles of CO₂ are produced from 1.00 g of baking powder?

(b) If 1 mol of CO₂ occupies 37.0 L at 350_F (a typical baking temperature), what volume of CO₂is produced from 1.00 g of baking powder?

Step-by-Step Solution

Answer

      a. The moles of CO₂ produced from 1.00 g of baking powder is 0.003 mol. 

      b.The volume of CO₂ produced from 1.00 g of baking powder is 0.111 L.

Step 1: Determination of moles of CO 2

The balanced equation for the given reaction is,

$\mathbf{Ca}{\left(H_2{\mathrm{PO}}_4\right)}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{NaHCO}}_{\mathbf{3}}\left(s\right)\mathbf{\to }\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}{\mathbf{CaHPO}}_{\mathbf{4}}\left(s\right)\mathbf{+}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}\left(s\right)$ 

 

From the above reaction, it can be concluded that 1 mol of Ca(H₂PO₄)₂ react with 2 moles of NaHCO₃ to produce 2 moles of CO₂.

 

Mass of baking powder = 1.00 g

 

Mass of Ca(H₂PO₄)₂ is,

$\mathbf{Mass}\mathbf{=}\frac{\mathbf{35}}{\mathbf{100}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{g}$ 

 

Mass of NaHCO₃ is,

 

 

Now, moles of Ca(H₂PO₄)₂ and NaHCO₃ are,

$$\begin{gathered} \mathbf{Ca}{\left(H_2{\mathrm{PO}}_4\right)}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{ }\mathbf{mass}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{g}}{\mathbf{234}\mathbf{.}\mathbf{05}\mathbf{g}\mathbf{/}\mathbf{mol}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0015}\mathbf{mol} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{NaHCO}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{ }\mathbf{mass}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{31}\mathbf{g}}{\mathbf{84}\mathbf{g}\mathbf{/}\mathbf{mol}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0037}\mathbf{mol} \end{gathered}$$ 

 

The limiting reagent in the reaction is Ca(H₂PO₄)₂.

 

1 mol of Ca(H₂PO₄)₂ produces 2 moles of CO₂. So, moles of CO₂ produced by 0.0015 mol of Ca(H₂PO₄)₂ are,

 $\mathbf{Moles}\mathbf{=}\mathbf{2}\mathbf{mol}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0015}\mathbf{mol}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{003}\mathbf{mol}$

 

Thus, the moles of CO₂ are produced from 1.00 g of baking powder are 0.003 mol.

Step 2: Determination of volume of CO 2

The volume occupied by 0.003 mol of CO₂ is,

 

$$\begin{gathered} \frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{n}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{2}}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{1}}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{L}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{003}\mathbf{mol}}{\mathbf{1}\mathbf{mol}} \\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{111}\mathbf{L} \end{gathered}$$ 

 

Thus, the volume of CO₂ produced from 1.00 g of baking powder is 0.111 L.