31.3 gram mixture NaBr and NaCl treated with H2SO4. 28.4 gram of Na2SO4 is produced. Then calculate the amount of NaCl and NaBr in the mixture.

Asked by Anil | 13th May, 2017, 12:32: PM

Expert Answer:

Moles of Na in Na2SO4

= 28.4/142  x  2 mol of Na / 1 mol Na2SO4 = 0.4

Weight of NaCl = x g

Weight of NaBr = 31.3 – x g

Number of moles of Na in NaCl  in the mixture is :

x / (58.4430 g NaCl/mol) x (1 mol Na / 1 mol NaCl) = x / 58.4430 

Number of moles of Na in NaBr in the mixture is :
(31.3 - x) / (102.8938 g NaBr/mol) x (1 mol Na / 1 mol NaBr) = (31.3 - x) / 102.8938 

Add the two expressions and set the sum equal to the total number of moles of Na found above: 
(x / 58.4430) + ((31.3 - x) / 102.8938) = 0.4

Solve for x algebraically: 
x = 12.9 g NaCl 
31.3 - 12.9 = 18.4 g NaBr

Answered by Prachi Sawant | 14th May, 2017, 12:33: AM