CBSE Class 11-science Answered
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.
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