What volume of chlorine gas at 45.3oC is needed to react with 14.2g of sodium to form NaCl at 1.72atm?

9.4L

Explanation:

In this particular question, we will need to write a balanced chemical equation for the reaction between sodium and chlorine to form sodium chloride.

Na + Cl - -> NaCl

Hence, we can see that 1 mole of chlorine reacts with 1 mole of sodium.

Now, we need to find the exact number of moles of chlorine atom that reacted with 14.2g of sodium. To do this, we simply divide the mass of the sodium by the atomic mass of the sodium which is 23.

Hence, the mass of sodium reacted is 14.2/23 which equals 0.617 moles

Simply because we have the mole ratio to be 1 to 1, it can be deduced that the number of moles of sodium reacted is also 0.617moles

Now, to get the volume of chlorine, we can use the ideal gas equation.

This is:

PV = nRT

V = nRT/P

Given:

v = ?

n = number of moles = 0.617 moles in this case

T = temperature = 45.3 + 273.15 = 318.45K

P = Pressure = 1.72 atm = 1.72 * 101325 pa = 174,279 Pa

R = molar gas constant = 8314.462L. Pa/K. mol

Inserting all these into the equation will yield:

V = (0.617 * 8314.462 * 318.45) / 174,279

V = 9.4L