If one-fourth cup of vinegar contains 0.049 mole of HC2H3O2, how many grams of baking soda will react with this amount of acetic acid? What would be the percent by weight of HC2H3O2 in this one-fourth cup of vinegar solution? (assume density of vinegar is 1.00 g/mL and that ¼ cup = 62.5 mL)

To determine the number of grams of baking soda that will react with 0.049 moles of acetic acid (HC2H3O2), we need to use the balanced chemical equation for the reaction between acetic acid and baking soda:

HC2H3O2 + NaHCO3 -> NaC2H3O2 + H2O + CO2

From the balanced equation, we can see that the stoichiometry is 1:1 between acetic acid and baking soda. This means that 1 mole of acetic acid reacts with 1 mole of baking soda.

Therefore, the number of grams of baking soda that will react with 0.049 moles of acetic acid can be calculated using the molar mass of baking soda (NaHCO3), which is approximately 84.01 g/mol.

Number of grams of baking soda = 0.049 moles * 84.01 g/mol = 4.12199 g

So, approximately 4.12 grams of baking soda will react with 0.049 moles of acetic acid.

Now, let's calculate the percent by weight of HC2H3O2 in the one-fourth cup (62.5 mL) of vinegar solution.

First, we need to calculate the mass of the vinegar solution using its density.

Mass of vinegar solution = density of vinegar * volume of vinegar
= 1.00 g/mL * 62.5 mL
= 62.5 g

Next, we need to determine the mass of HC2H3O2 in the vinegar solution using the percent composition of acetic acid, which is approximately 4-5%.

Mass of HC2H3O2 in vinegar solution = Percent composition of HC2H3O2 * Mass of vinegar solution / 100
= 4-5% * 62.5 g / 100
= 2.5 g - 3.125 g

So, the percent by weight of HC2H3O2 in the one-fourth cup of vinegar solution is approximately 2.5% - 3.125%.

To find out how many grams of baking soda will react with the given amount of acetic acid, we need to use balanced chemical equation for the reaction between acetic acid (HC2H3O2) and baking soda (sodium bicarbonate, NaHCO3). The balanced chemical equation is:

HC2H3O2 + NaHCO3 -> NaC2H3O2 + H2O + CO2

From the equation, we can see that the molar ratio between HC2H3O2 and NaHCO3 is 1:1. This means that 1 mole of HC2H3O2 reacts with 1 mole of NaHCO3.

Given that 0.049 moles of HC2H3O2 are present, we can conclude that the same amount of moles of NaHCO3 will react with HC2H3O2.

Now, the molar mass of HC2H3O2 is 60.05 g/mol. Therefore, 0.049 moles of HC2H3O2 would weigh:

0.049 moles x 60.05 g/mol = 2.94 g

So, 2.94 grams of baking soda will react with the given amount of acetic acid.

To calculate the percent by weight of HC2H3O2 in the vinegar solution, we need to find the weight of HC2H3O2 and divide it by the weight of the solution, then multiply by 100.

Given that the vinegar has a density of 1.00 g/mL, ¼ cup (62.5 mL) of vinegar would weigh:

62.5 mL x 1.00 g/mL = 62.5 g

We already calculated that the weight of HC2H3O2 is 2.94 g.

Therefore, the percent by weight of HC2H3O2 in the one-fourth cup of vinegar solution would be:

(2.94 g / 62.5 g) x 100 = 4.70%

CXC