13.4 g hydroiodic acid 46.6 mL 0.374 M barium hydroxide percent by mass

13.4 g hydroiodic acid 46.6 mL 0.374 M barium hydroxide percent by mass

 

 

A 13.4 g sample of an aqueous solution of hydroiodic acid

contains an unknown amount of the acid.

If 46.6 mL of 0.374 M barium hydroxide are required

to neutralize the hydroiodic acid,

what is the percent by mass of hydroiodic acid in the mixture?

 

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반응한 Ba(OH)2의 몰수를 계산하면,

(0.374 mol/L) (46.6/1000 L) = 0.0174284 mol Ba(OH)2

 

 

 

2HI + Ba(OH)2 → BaI2 + 2H2O

HI : Ba(OH)2 = 2 : 1 계수비(= 몰수비) 이므로,

반응한 HI의 몰수를 계산하면,

HI : Ba(OH)2 = 2 : 1 = ? mol : 0.0174284 mol

 

? = 2 × 0.0174284 = 0.0348568 mol HI

 

 

 

HI의 몰질량 = 127.91 g/mol 이므로,

0.0348568 mol × (127.91 g/mol) = 4.4585 g HI

 

 

 

질량 백분율을 계산하면,

(4.4585 g / 13.4 g) × 100 = 33.3%

 

 

 

답: 33.3%