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일반화학/[03장] 화학량론

0.1058 g H2C2O4 + 28.97 mL MnO4^-

by 영원파란 2020. 8. 26.

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0.1058 g H2C2O4 + 28.97 mL MnO4^-

 

 

A solution of permanganate is standardized

by titration with oxalic acid (H2C2O4).

It required 28.97 mL of the permanganate solution

to react completely with 0.1058 g of oxalic acid.

The unbalanced equation for the reaction is

MnO4^-(aq) + H2C2O4(aq) → Mn^2+(aq) + CO2(g)

What is the molarity of the permanganate solution?

 

---------------------------------------------------

 

H2C2O4의 몰질량 = 90.03 g/mol 이므로,

0.1058 g / (90.03 g/mol) = 0.001175 mol H2C2O4

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

균형 맞춘 반응식

2MnO4^- + 5H2C2O4 + 6H^+ → 2Mn^2+ + 10CO2 + 8H2O

( 참고 https://ywpop.tistory.com/6357 )

 

MnO4^- : H2C2O4 = 2 : 5 계수비(= 몰수비) 이므로,

0.001175 mol H2C2O4와 반응한 MnO4^-의 몰수를 계산하면,

MnO4^- : H2C2O4 = 2 : 5 = ? mol : 0.001175 mol

? = 2 × 0.001175 / 5 = 0.0004700 mol MnO4^-

 

 

 

MnO4^-의 몰농도를 계산하면,

0.0004700 mol / (28.97/1000 L) = 0.01622 mol/L

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

답: 0.01622 M

 

 

 

 

[키워드] redox MnO4^- + H2C2O4, redox H2C2O4 + MnO4^-, 0.1058 g 28.97 mL MnO4^-(aq) + H2C2O4(aq), 옥살산 0.1058 g 과망가니즈산 염 용액 28.97 mL

 

 

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