Question

1g of a mixture NaCl and NaBr are reacted with H2SO4 and whole is converted into Na2SO4.The weight of Na2SO4 is found to be 2g. Find out the % of NaCl and NaBr in the mixture..
Plz answer this question..

Answer 1

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 - xg 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) / 1028938 g NaBr/mol) x (1 mol Na/ 1 mol NaBr) = (31.3 - x) / 1028938 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) / 1028938 = 0.4 Solve for x algebraically:

X 12.9 g NaCl

31.3 - 12.9 = 18.4 g NaBr

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