Question

consider the following series of reaction (1) Cl2+2NaOH =NaCl+NaClO+H2O​

Answer 1

Answer:

Factor label method: (MwofNaClO4=122.5MwofNaClO3=106.5gmol−1)

xg of Cl2=⎛⎝⎜⎜⎜⎜122.5gNaClO4122.5gNaClO4per mol ofNaClO4⎞⎠⎟⎟⎟⎟(4molNaClO32molNaClO4)

(3molNaClO1molNaClO3)(1molCl21molNaClO)(71.0gCl21molCl2)

=122.5122.5×43×31×11×71.01=284.0g

Mole method:

n(NaClO)=n(Cl2),

n(NaClO3)=13n(NaClO)=13(Cl2)

n(NaClO4)=34n(NaClO3)=34×13n(Cl2)=14n(Cl2)

n(NaClO4)=122.5gNaClO4122.5ofNaClO4/moleNaClO4

=1.0molNaClO4

n(Cl2)=4×1.0=4molCl2

Mass of (Cl2)=4×71.0g=284gCl2