4.80 dm^3 of chlorine gas was reacted with NaOH solution.

a) how many moles of Cl2 reacted
b) what mass of NaOCl was formed?
c) if the concentration of the NaOH was 2.00 mol/dm^3. What volume of sodium hydroxide solution was required?

d) write an ionic equation for this reaction

a) Well, let's do some math tricks! 1 mole of any gas occupies 22.4 dm^3 at room temperature and pressure. So, if we have 4.80 dm^3 of Chlorine gas, we can simply divide it by 22.4 to find the number of moles. That's approximately 0.214 moles. Voila!

b) To find the mass of NaOCl formed, we need to review our stoichiometry skills. First, we need to balance the equation for the reaction between chlorine gas (Cl2) and NaOH. It goes like this: Cl2 + 2NaOH -> NaOCl + NaCl + H2O. According to the equation, we have a 1:1 molar ratio between Cl2 and NaOCl. Since we have 0.214 moles of Cl2, that means we'll also have 0.214 moles of NaOCl. The molar mass of NaOCl is approximately 74.5 g/mol, so multiplying it by 0.214 will give us around 15.9 grams of NaOCl. Easy peasy, lemon squeezy!

c) Now, let's calculate the volume of sodium hydroxide solution required. We know that the concentration of NaOH is 2.00 mol/dm^3, which means for every 1 dm^3 of solution, we have 2.00 moles of NaOH. Since we need to find the volume required for 0.214 moles of NaOH, we can use a little proportionality: 0.214 moles is to X dm^3, as 2.00 moles is to 1 dm^3. Cross multiplying, we get X ≈ 0.107 dm^3. So, approximately 0.107 dm^3 or 107 mL of NaOH solution is needed. Bottoms up!

d) Now, behold the mighty ionic equation for this reaction: Cl2 + 2NaOH -> NaOCl + NaCl + H2O. In this equation, we showcase the glory of chloride ions (Cl-) joining forces with sodium hydroxide (NaOH) to create sodium hypochlorite (NaOCl), sodium chloride (NaCl), and everyone's favorite, water (H2O). Bravo!

a) To determine the number of moles of Cl2 that reacted, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas (assuming it is constant)
V = volume of the gas (4.80 dm^3)
n = number of moles of the gas (what we are trying to find)
R = ideal gas constant (8.314 J/(mol·K))
T = temperature of the gas (assuming it is constant)

Since pressure and temperature are assumed constant, we can rearrange the equation to solve for moles:

n = (PV) / RT

Given the values:
P = constant
V = 4.80 dm^3
R = 8.314 J/(mol·K)
T = constant

Substituting the values, we can calculate the number of moles of Cl2 that reacted.

b) To determine the mass of NaOCl formed, we need to use stoichiometry and the balanced equation for the reaction. Unfortunately, you haven't provided the balanced equation for the reaction between Cl2 and NaOH. Please provide the balanced equation so we can proceed with this calculation.

c) To determine the volume of sodium hydroxide solution required, we need to use the equation:

C = n/V

Where:
C = concentration of the solution (2.00 mol/dm^3)
n = number of moles of solute (what we want to find)
V = volume of the solution in dm^3 (what we want to find)

Rearranging the equation, we have:

n = C × V

Given the concentration of NaOH (2.00 mol/dm^3), we can substitute the values and calculate the volume of sodium hydroxide solution required.

d) Once you provide the balanced equation for the reaction, I can help you write the ionic equation.

a) To find the number of moles of Cl2 that reacted, you need to use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. However, since you haven't provided the pressure, temperature, or ideal gas constant values, we'll assume standard conditions (P = 1 atm, T = 273.15 K, R = 0.0821 L.atm/mol.K).

Given:
Volume of chlorine gas (V) = 4.8 dm^3

Using the equation PV = nRT, we can rearrange it and solve for n:
n = PV / RT

Considering the standard conditions:
n = (1 atm) * (4.8 dm^3) / (0.0821 L.atm/mol.K * 273.15 K)

Calculating this expression gives you the number of moles of Cl2 that reacted.

b) To determine the mass of NaOCl formed, you need to know the balanced chemical equation for the reaction between chlorine gas (Cl2) and sodium hydroxide (NaOH). Unfortunately, you haven't provided this information, so we can't solve this specific question without it.

c) To find the volume of sodium hydroxide (NaOH) solution required, you need to know the number of moles of NaOH and the concentration of the NaOH solution. The concentration of NaOH solution is given as 2.00 mol/dm^3.

Given:
Desired concentration (C) = 2.00 mol/dm^3

Using the formula C = n / V, where n is the number of moles and V is the volume in liters, we can rearrange the equation and solve for V:
V = n / C

Using the number of moles of NaOH and the concentration, you can calculate the required volume of the NaOH solution.

d) To write the ionic equation for this reaction, we need the balanced chemical equation. Since it wasn't provided, it's not possible to accurately write the ionic equation. However, a possible reaction could be:

Cl2 + 2NaOH -> NaOCl + NaCl + H2O

This equation suggests the formation of sodium hypochlorite (NaOCl), sodium chloride (NaCl), and water (H2O) from the reaction between chlorine gas (Cl2) and sodium hydroxide (NaOH). Note that without the balanced equation, we cannot guarantee the accuracy of this ionic equation.

a. Actually, no one knows; however, if all 4.80 dm^3 reacted AND the Cl2 gas was at STP, then it is 4.80/22.4 = ?

b. Can't answer b,c,d. Is NaOH concd or dilute? Is it hot or cold?
Here are a couple of reactions but there are others.
2 NaOH (conc., cold) + Cl2 = NaClO + NaCl + H2O

6 NaOH (conc.,hot) + 3Cl2 = NaClO3 + 5NaCl + 3H2O.