Question c4615

Apr 16, 2015

Sodium bicarbonate, $N a H C {O}_{3}$, and acetic acid, $C {H}_{3} C O O H$, will react to produce sodium acetate, $C {H}_{3} C O O N a$, and carbonic acid, ${H}_{2} C {O}_{3}$.

The carbonic acid will then decompose to give water and carbon dioxide, $C {O}_{2}$.

NaHCO_(3(s)) + CH_3COOH_((aq)) -> CH_3COONa_((aq)) + underbrace(H_2O_((l)) + CO_(2(g)))_(color(blue)(H_2CO_3)

So, you've got 8.4 g of sodium bicarbonate reacting with 20g of acetic acid, and producing 4.4 g of carbon dioxide. The first thing you need is work backwards from the mass of $C {O}_{2}$ produced to determine what is the mass of acetic acid that reacted.

To do this, use the compounds' molar masses to calculate how many moles of each you have

4.4cancel("g") * ("1 mole "CO_2)/(44.01cancel("g")) ~= "0.10 moles " $C {O}_{2}$

8.4cancel("g") * ("1 mole "NaHCO_3)/(84.007cancel("g")) ~= "0.10 moles " $N a H C {O}_{3}$

According to the $1 : 1$ mole ratio that exists between sodium bicarbonate, acetic acid, and carbon dioxide, the number of moles of acetic acid that reacted is

0.10cancel("moles "CO_2) * ("1 mole "CH_3COOH)/(1cancel("mole "CO_2)) = "0.10 moles " $C {H}_{3} C O O H$

Therefore, the mass that reacted was

0.10cancel("moles "CH_3COOH) * "60.005 g"/(1cancel("mole " CH_3COOH)) ~= "6.0 g " $C {H}_{3} C O O H$

Therefore, the total mass of the reactants wil be

${m}_{\text{reactants" = 8.4 + 6.0 = "14.4 g}}$

According to the law of mass conservation, the total mass of the products must be equal to the total mass of the reactants., which means that

${m}_{\text{products" = m_"reactants" = "14.4 g}}$

Therefore,

${m}_{\text{products" = m_(CO_2) + m_"residue}}$, or

${m}_{\text{residue" = m_"products" - m_(CO_2) = 14.4 - 4.4 = "10.0 g}}$

${\overbrace{N a H C {O}_{3 \left(s\right)}}}^{\textcolor{b l u e}{\text{8.4 g")) + overbrace(CH_3COOH_((aq)))^(color(blue)("6.0 g")) -> overbrace(CH_3COONa_((aq)) + H_2O_((l)))^(color(green)("10.0 g")) + overbrace(CO_(2(g)))^(color(green)("4.4 g}}}$

Apr 16, 2015

The mass of the residue, or sodium acetate solution, should be $\text{24.0 g}$ of sodium acetate solution.

In this reaction, sodium bicarbonate reacts with acetic acid to produce sodium acetate, carbon dioxide, and water. The carbon dioxide bubbles out of solution, leaving behind a sodium acetate solution.

$\text{NaHCO"_3 + "CH"_3"COOH}$$\rightarrow$$\text{CH"_3"COONa + CO"_2 + "H"_2"O}$

According to the law of conservation of mass, the total mass of the reactants equals the total mass of the products. The reactants are 8.4 g sodium bicarbonate and 20 g acetic acid solution. The total mass of the reactants is:

$\text{8.4 g NaHCO"_3 + "20 g CH"_3"COOH}$ = $\text{28.4 g total mass of reactants}$

The total mass of the products must equal the mass of the reactants. If we subtract the mass of carbon dioxide from the total mass of the reactants, we can determine the mass of the residue, which is basically a sodium acetate solution.

$\text{28.4 g total mass of reactants - 4.4g CO"_2}$ = "24.0 g of residue" ("sodium acetate solution")#