

InterviewSolution
Saved Bookmarks
1. |
A solid mixture `5 g` consists of lead nitrate and sodium nitrate was heated below `600^(@)C` until weight of residue was constant. If the loss in weight is `28%` find the amount of lead nitrate and sodium nitrate in mixture. |
Answer» Mass of solid mixture = 5.0 g Let the mass of lead nitrate = x g `:.` Mass of sodium nitrate `= (5-x)g` Loss in mass of mixture `= (5xx28)/(100)=1.4g` `:.` Mass of the residue `= 5 - 4 = 3.6 g` Calculation of the mass of the residue from lead nitrate. `underset(underset("= 663 g")(2(207+28+96)))(2Pb(NO_(3))_(2))rarrunderset(underset(=446g)(2(207+16)))(2PbO)+4NO_(2)+O_(2)` 662 g of `Pb(NO_(3))_(2)` form residue (PbO) = 446 g `:. x` g of `Pb(NO_(3))_(2)` form residue `(PbO)=(446)/(662) xx xg` Step II. Calculation of the mass of the residue from sodium nitrate `underset(underset(=170g)(2(23+14+48)))(2NaNO_(3))rarrunderset(underset(=138g)(2(23+14+32)))(2NaNO_(2)+O_(2))` 170g of `NaNO_(3)` form residue `(NaNO_(2))=138g` `:. (5-x)g` of `NaNO_(3)` form residue `(NaNO_(2))=(138)/(170) xx (5-x)g` Step III. Calculation of the amount of lead nitrate and sodium nitrate Total mass of PbO and `NaNO_(3)=[(446)/(662)x+(138)/(170)(5-x)]g` Total mass of the residue actually left = 3.6 g `:. (446)/(662)x+(138)/(170)(5-x)=3.6` `0.674x+0.812(5-x)=3.6` or `0.674x+4.06 -0.812x = 3.6` or `0.674x-0.812x = 3.6 - 4.06` `-0.13x = - 0.46` `:. x = (0.46)/(0.138)=3.38g` Thus, Mass of `Pb(NO_(3))_(2)=3.38g` Mass of `NaNO_(3)=5 - 3.38 = 1.62g`. |
|