How many grams of NaCl was used by the electrolysis method to obtain 74L of chlorine gas at a temperature of 27 degrees C at 1 atm?

351g

Explanation:

The following were obtained from the question:

Volume (V) of Cl2 = 74L

Temperature (T) = 27°C = 27°C + 273 = 300K

Pressure (P) = 1 atm

To obtain the mass of NaCl used in the reaction, we need to determine the number of mole Cl2 produced from the reaction. This is illustrated below:

Using the ideal gas equation, the number of mole of Cl2 can be obtained as follow:

Volume (V) of Cl2 = 74L

Temperature (T) = 300K

Pressure (P) = 1 atm

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) of Cl2 =?

PV = nRT

n = PV /RT

n = (1 x 74) / (0.082 x 300)

n = 3 moles

Therefore, the reaction produced 3 moles of Cl2.

Next, the balanced equation for the reaction. This is given below:

2NaCl —> 2Na + Cl2

From the balanced equation above,

2 moles of NaCl produced 1 mole of Cl2.

Therefore, Xmol of NaCl will produce 3 moles of Cl2 i.e

Xmol of NaCl = 2 x 3

Xmol of NaCl = 6 moles

Therefore, 6 moles of NaCl was used in the reaction.

Next, we shall convert 6 mole of NaCl to grams. This is illustrated below:

Number of mole NaCl = 6 moles

Molar Mass of navl= 23+ 35.5 = 58.5g/mol

Mass of NaCl =?

Mass = number of mole x molar Mass

Mass of NaCl = 6 x 58.5

Mass of NaCl = 351g

From the calculations made above, 351g of NaCl was used in the electrolysis.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g