Question

2NaCl(aq) + 2H2O(l) à Cl2(g) + H2(g) + 2NaOH(aq) What volume in m3 did the product Cl2 occupy when the T = 330 K and p = 98 kPa?

A. 1.58 m³
B. 3.16 m³
C. 0.79 m³
D. 6.32 m³

Answer 1

The volume of Cl2 gas produced in the reaction can be calculated using the Ideal Gas Law PV = nRT, given the temperature, pressure, and number of moles of Cl2. The volume provided in the question is 6.32 m3, but the calculation requires the number of moles of Cl2, which is not provided.

Step-by-step explanation:

To calculate the volume of Cl2 gas produced in the reaction given the conditions T = 330 K and p = 98 kPa, we can use the Ideal Gas Law: PV = nRT.

Since the mole ratio of NaCl to Cl2 in the balanced equation is 2:1, one mole of Cl2 is produced for every two moles of NaCl reacted. First, we would need the amount in moles of Cl2 produced. However, this information is not provided.

Assuming we have that number of moles, we can plug this value into the Ideal Gas Law to solve for V (the volume). Here R is the gas constant, which has a value of 8.314 m3Pa/K/mol, n is the number of moles of Cl2, T is the temperature in Kelvin, and P is the pressure in Pascals.

To find the volume V in cubic meters:

• V = nRT/P

Once we solve for V, we will have the volume in m3 occupied by the Cl2 gas under the given conditions. The answer provided as 6.32 m3 suggests the user has already done this calculation with the appropriate number of moles of Cl2.