2) One way of making sodium chloride is according to the following unbalanced equation: Na(s) + Cl2(g) → NaCl(s) In a reaction, a student mixed 3.00 grams of Na(s) with 6.2 grams of Cl2(g), making 7.60 grams of NaCl. Balance the equation and determine which, if any, of the starting materials was in excess? . [Molar mass (g/mol): Na = 22.99, and Cl = 35.45] 3) If 27.0 mL of 0.25 M NaCl aqueous solution are mixed with 36.0 mL of 0.42 M Ca(NO3)2 aqueous solution. Is there a reaction taking place? Find the molarity of each ion present in the final solution. 4) Vinegar contains acetic acid (HC2H3O2), which is responsible for its acidity. In one analysis, a 20.00 mL sample of vinegar was titrated with 0.50 M NaOH(aq). It required 35.5 mL of this sodium hydroxide titrant to neutralize the acid in the vinegar sample. If 1.000 liter of this vinegar weighs 1.008 kg, what is the percent, by weight, of acetic acid in this vinegar sample? . [Molar mass (g/mol): C = 12.01, H = 1.008, Na = 22.99, and O = 16] 5) A 50.0 mL sample of 18.0 M sulfuric acid was diluted with enough water in a volumetric flask to make 5.00x10^2 mL of solution. A 50.0 mL aliquot of this solution was then further diluted to a volume of 2.50x10^2 mL. What is the molarity of the solution after the second dilution?

2) To balance the equation, we need to make sure that the number of atoms of each element is equal on both sides of the equation.

The equation without balancing is:
Na(s) + Cl2(g) → NaCl(s)

To balance the equation, we add coefficients in front of the reactants and products:
2 Na(s) + Cl2(g) → 2 NaCl(s)

Now, let's calculate the number of moles of each substance.

Number of moles of Na(s) = mass / molar mass = 3.00 g / 22.99 g/mol ≈ 0.1306 mol
Number of moles of Cl2(g) = mass / molar mass = 6.2 g / 35.45 g/mol ≈ 0.1747 mol

From the balanced equation, we can see that the mole ratio between Na(s) and Cl2(g) is 2:1.
Therefore, the theoretical amount of Na(s) needed to react with 0.1747 mol of Cl2(g) is 0.1747 mol / 2 = 0.0873 mol.

Since we have 0.1306 mol of Na(s) available, which is greater than the theoretical amount needed (0.0873 mol), Na(s) is in excess.

Answer: Na(s) was in excess.

3) To determine if a reaction takes place, we need to check if there is a possible chemical reaction between NaCl and Ca(NO3)2. By looking at the formulas, we can see that both compounds contain ions (Na+, Cl-, Ca2+, NO3-) and can potentially react.

To find the molarity of each ion present in the final solution, we need to calculate the moles of each ion.

For NaCl:
Number of moles of NaCl = volume(mL) * molarity = 27.0 mL * 0.25 mol/L = 6.75 mmol
The concentration of Na+ and Cl- ions in the final solution is 0.25 M.

For Ca(NO3)2:
Number of moles of Ca(NO3)2 = volume(mL) * molarity = 36.0 mL * 0.42 mol/L = 15.12 mmol
Since Ca(NO3)2 dissociates into Ca2+ and 2 NO3- ions, the molar concentration of Ca2+ ions is 0.42 M, while the molar concentration of NO3- ions is 2 * 0.42 M = 0.84 M.

Answer: The molarity of each ion present in the final solution is:
- Na+: 0.25 M
- Cl-: 0.25 M
- Ca2+: 0.42 M
- NO3-: 0.84 M

4) To find the percent, by weight, of acetic acid in the vinegar sample, we need to determine the moles of acetic acid and the weight percent.

The equation for the reaction between acetic acid and NaOH is:
HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l)

From the balanced equation, we can see that the mole ratio between acetic acid and NaOH is 1:1.
Therefore, the moles of acetic acid in the vinegar sample is equal to the moles of NaOH used in the titration.

Number of moles of NaOH = volume(L) * molarity = 0.0355 L * 0.50 mol/L = 0.0178 mol

Since the molar mass of acetic acid (HC2H3O2) = 12.01 g/mol + 2 * 1.008 g/mol + 3 * 16.00 g/mol = 60.05 g/mol, we can calculate the weight of acetic acid in the vinegar sample.

Weight of acetic acid = moles * molar mass = 0.0178 mol * 60.05 g/mol = 1.067 g

To calculate the percent, by weight, of acetic acid in the vinegar sample, we use the following formula:

Percent by weight = (weight of acetic acid / weight of vinegar sample) * 100

The weight of vinegar sample = 1.008 kg = 1008 g

Percent by weight = (1.067 g / 1008 g) * 100 ≈ 0.1059%

Answer: The percent, by weight, of acetic acid in the vinegar sample is approximately 0.1059%.

5) To find the molarity of the solution after the second dilution, we need to use the concept of dilution.

The formula for dilution is:
C1V1 = C2V2

Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume

Using the given data:
C1 = 18.0 M
V1 = 50.0 mL = 0.0500 L
V2 = 2.50x10^2 mL = 0.250 L

Rearranging the formula, we can solve for C2:

C2 = (C1V1) / V2
C2 = (18.0 M * 0.0500 L) / 0.250 L
C2 = 3.6 M

Answer: The molarity of the solution after the second dilution is 3.6 M.