Answer to Question #269290 in General Chemistry for Amna

Question

based on the following reaction cl2+Naclo=NaCL+CLO2 ,

2.5 MOCL2 measured at 20 c and 4.12 atm is reacted with 800 mL of 3.30 M NaCLO F 51.8 g of ClO2 is obtained what is the percent yield Of the reaction determine the limiting and dream meaning as gram reagent

Answer 1

​=122.5gmol−1;Mw of NaClO3​=106.5gmol−1)
x g of Cl2​=(122.5 g NaClO4​ permole of NaClO3​
122.5 g NaClO4​​)(3 mol NaClO4​
4 mol NaClO3​​)(1 mol NaClO3​
3 molNaClO​)(1 mol NaClO
1 mol Cl2​​)                                                                                                                                                                   ×(1.0 mol Cl2​
71.0 g Cl2​​)
x g of Cl2​=122.5
122.5​×3
4​×1
3​×1
1​×1
71.0​
Mole Method : n(NaClO)=n(Cl2​)
n(NaClO2​)=3
1​n(NaClO)=3
1​n(Cl2​)
n(NaClO4​)=4
3​n(NaClO3​)=4
3​×3
1​n(Cl2​)=4
1​n(Cl2​)
n(NaClO4​)=122.5 g NaClO4​
122.5 g NaClO4​​=1.0 mol NaClO4​
n(Cl2​)=4×1.0=4 mol Cl2​
Mass of(Cl2​)=4×71.0g=284.0gCl2limiting reagent : Sodium Gram reagent :chlorine dioxide