Question

d. another way of writing the chemical formula of vinegar is ch₃cooh. use ionic naming rules to write the name of this chemical compound (1 point for correct elements written, 1 point for correct polyatomic, 1 point for correct order of elements)
part 4: reactions and molar mass
when dropped in water (h₂o), sodium (na) will react violently producing hydrogen gas (h₂) and the base lye/sodium hydroxide (naoh)
a. write out the chemical equation for this reaction (you dont need to balance it here) (3pts: 1pt for proper reaction equation setup, 1pt for products being on the correct side of the arrow, 1pt for reactants being on the correct side of the arrow)
b. rewrite your chemical equation and balance it so there are equal amounts of each element on both sides of the equation (2 pts: 0.5 pt for each compound properly balanced)
c. calculate the molar mass of naoh, show your work (1pt: 0.5pts for correct molar mass, 0.5pts for showing your work)

Explanation:

Step1: Name $\text{CH}_3\text{COOH}$

This is a carboxylic acid; the polyatomic ion $\text{CH}_3\text{COO}^-$ is acetate, so the compound is acetic acid (following ionic naming for the carboxylate and hydrogen cation).

Step2: Write unbalanced reaction

Identify reactants (Na, $\text{H}_2\text{O}$) and products ($\text{H}_2$, NaOH), place on correct sides.
$\text{Na} + \text{H}_2\text{O}
ightarrow \text{NaOH} + \text{H}_2$

Step3: Balance the chemical equation

Adjust coefficients to equalize element counts: 2 Na, 2 $\text{H}_2\text{O}$ on reactant side; 2 NaOH, 1 $\text{H}_2$ on product side.
$2\text{Na} + 2\text{H}_2\text{O}
ightarrow 2\text{NaOH} + \text{H}_2$

Step4: Calculate molar mass of NaOH

Sum atomic masses: $\text{Na}=22.99\ \text{g/mol}$, $\text{O}=16.00\ \text{g/mol}$, $\text{H}=1.01\ \text{g/mol}$.
$\text{Molar Mass} = 22.99 + 16.00 + 1.01$

Answer:

d. Acetic acid
a. $\text{Na} + \text{H}_2\text{O}
ightarrow \text{NaOH} + \text{H}_2$
b. $2\text{Na} + 2\text{H}_2\text{O}
ightarrow 2\text{NaOH} + \text{H}_2$
c. $40.00\ \text{g/mol}$