Question

c₂h₄o₂ + nacho₃ ⟶ co₂ + nac₂h₃o₂ + h₂o
vinegar baking soda carbon dioxide sodium acetate water
products reactants
__total na atoms __total na atoms
__total o atoms __total o atoms
__total c atoms __total c atoms
__total h atoms __total h atoms
__total of all atoms __total of all atoms
was matter conserved? explain:

Explanation:

Step1: Balance the chemical equation

First, balance the given unbalanced equation:
Unbalanced: $\text{C}_2\text{H}_4\text{O}_2 + \text{NaCHO}_3
ightarrow \text{CO}_2 + \text{NaC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O}$
Balanced: $\text{C}_2\text{H}_4\text{O}_2 + \text{NaCHO}_3 = \text{CO}_2 + \text{NaC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O}$
(This equation is already balanced when counting each atom type.)

Step2: Count reactant atoms

Na atoms: 1 (from $\text{NaCHO}_3$)

O atoms: $2 + 3 = 5$ (2 from $\text{C}_2\text{H}_4\text{O}_2$, 3 from $\text{NaCHO}_3$)

C atoms: $2 + 1 = 3$ (2 from $\text{C}_2\text{H}_4\text{O}_2$, 1 from $\text{NaCHO}_3$)

H atoms: $4 + 1 = 5$ (4 from $\text{C}_2\text{H}_4\text{O}_2$, 1 from $\text{NaCHO}_3$)

Total atoms: $1 + 5 + 3 + 5 = 14$

Step3: Count product atoms

Na atoms: 1 (from $\text{NaC}_2\text{H}_3\text{O}_2$)

O atoms: $2 + 2 + 1 = 5$ (2 from $\text{CO}_2$, 2 from $\text{NaC}_2\text{H}_3\text{O}_2$, 1 from $\text{H}_2\text{O}$)

C atoms: $1 + 2 = 3$ (1 from $\text{CO}_2$, 2 from $\text{NaC}_2\text{H}_3\text{O}_2$)

H atoms: $3 + 2 = 5$ (3 from $\text{NaC}_2\text{H}_3\text{O}_2$, 2 from $\text{H}_2\text{O}$)

Total atoms: $1 + 5 + 3 + 5 = 14$

Step4: Evaluate matter conservation

Compare total reactant and product atoms; they are equal.

Answer:

Products	Reactants
5 Total O atoms	5 Total O atoms
3 Total C atoms	3 Total C atoms
5 Total H atoms	5 Total H atoms
14 Total of all Atoms	14 Total of all Atoms

Was matter conserved? Explain: Yes, matter was conserved. The total number of each type of atom (and total atoms overall) is identical in the reactants and products, which follows the law of conservation of mass.