Answer:
351g
Step-by-step 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.