Question

chm1110 laboratory
limiting reactant
there would be 0.502 grams of acetic acid

1. now convert the mass of acetic acid (from step 2) to moles of acetic acid. (5 points)

8.36 x 10^-3 moles
using the balanced chemical reaction between acetic acid and baking soda depicted below, please convert the moles of acetic acid (from step 3) to moles of sodium bicarbonate (5 points)
ch₃cooh(l) + nahco₃(s)→na⁺(aq)+ch₃coo⁻(aq) + co₂(g) + h₂o(l)

1. convert the moles of sodium bicarbonate (from step 4) to the mass of baking soda when there would be equal moles of acetic acid and sodium bicarbonate present (i.e. they would be stoichiometrically equivalent). (10 points)

Explanation:

Step1: Identify mole - ratio from balanced equation

From the balanced equation $CH_3COOH_{(l)}+NaHCO_3_{(s)}
ightarrow Na^+_{(aq)} + CH_3COO^-_{(aq)}+CO_2_{(g)}+H_2O_{(l)}$, the mole - ratio of $CH_3COOH$ to $NaHCO_3$ is 1:1.

Step2: Recall moles of acetic acid

The moles of acetic acid from step 3 is $n_{CH_3COOH}=8.36\times 10^{-3}$ moles. Since the mole - ratio of $CH_3COOH$ to $NaHCO_3$ is 1:1, the moles of $NaHCO_3$ is also $n_{NaHCO_3}=8.36\times 10^{-3}$ moles.

Step3: Calculate mass of $NaHCO_3$

The molar mass of $NaHCO_3$ is $M = 22.99+1.01 + 12.01+3\times16.00=84.01$ g/mol. Using the formula $m = n\times M$, we have $m_{NaHCO_3}=n_{NaHCO_3}\times M=(8.36\times 10^{-3}\text{ mol})\times84.01\text{ g/mol}\approx0.702$ g.

Answer:

0.702 g