A solid mixture 5 gram consists of lead nitrate and sodium nitrate was heated below 6000C until weight of residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in mixture.

Following are thermal decomposition  processes.PbNO32a g→PbO+2NO2↑+12O2↑NaNO3b g→NaNO2+12O2↑∴a+b=5                    .....1The loss in weight for 5 gram mixture  =5×28100=1.4 gramResidue left = 5 – 1.4 = 3.6 gramThe residue contains PbO+NaNO2∵331 gram PbNO32 gives =223 g PbO∴  a g PbNO32 gives =223×a 331g PbO∵  85 gram NaNO3 gives =69 gram NaNO2∴  b g NaNO3 gives =69×b85 gram NaNO2∴  223×a331+69×b85=3.6                             …..(2)Solving equations (1) and (2),∴  a=3.32 gram,  b=1.68 gram