Barium chloride + sodium sulphate --> barium sulphate + sodium chloride
BaCl2 + Na2SO4 ---> BaSO4 + 2 NaCl
The barium sulphate appears as a white precipitate
Silver nitrate + Sodium chloride ---> Silver Chloride + sodium nitrate
AgNO3 + NaCl ----> AgCl + NaNO3
The silver chloride appears as a white precipitate.
These are sometimes called double decomposition reactions.
In order to obtain solid NaCl, the student should do a few steps.
First, he/she should do filtration. Pass the mixture through a filter paper, where all the sand should be filtered out already because they're not dissolved in the solution plus they're too small to pass through the filter paper.
Next, the filtrate should be left with NaCl (aqueous state). To seperate NaCl with the liquid, the student can either do evaporation or crystallization, depending on how pure or fast he/she wants the results to be. Evaporation involves heating the beaker or whatever apparatus under the bunsen burner until all the liquid has evaporated. Then, some white powder should be left, they're NaCl solid. For crystallization, the student should just put the beaker on a room condition environment, and wait. They might have to wait a month or so for the liquid to completely evaporate itself and left with clear and pure NaCl crystals.
hydrogen and carbon, hope that helped
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
