Calculations on chemical equations and its solution with examples

 Calculations on chemical equations

Calculations on chemical equations (stoichiometric calculations) are based on the mass conservation law of substances. In real chemical processes because of incomplete running chemical reactions and losses a products mass usually less theoretically calculated.
Yield of reaction (h) is a ratio of real product mass (mr) to theoretically possible (mt), expressed in shares of units or percent.
If in conditions of problems a yield of reactions products does not specified, its take in calculations as 100% should be minded that;
 (quantitative output).

h=mr

––* 100%mt


example 

Calculated the mass of copper formed when reduction 8 g copper oxide by hydrogen. The yield of reaction – 82% from theoretical.

solution
CuO + H2 ® Cu + H2O


example :
Calculate the yield of reaction of tungsten preparation by aluminizing, if from 33.14 g of ore concentrate, containing WO3 and non-reduce admixtures (mass share of admixtures 0.3), were obtained 12.72 g of metal.

solution
1.      Define the WO3 mass (g) in 33.14 g of ore concentrate.

w(WO3) = 1.0 – 0.3 = 0.7
m(WO3) = w(WO3) * more = 0.7 * 33.14 = 23.2 g

2.      Define the theoretical yield of tungsten as a result of reduction 23.2 g by aluminum powder.

WO3 + 2Al ® Al2 + O3 + W

When reducing 232 g (1g-mol) WO3 form 187 g (1g-mol) W, but at 23.2 g WO3 – X g W

X =
23.2 * 187
—————
232
= 18.7 g W

3.      Calculate the practical yield of tungsten

18.7 g W
––
100%
12.72 g W
––
Y%

Y =
12.72 * 100
–––––––––
18.7
= 68 %

example;
How much grams of barium sulphate will be formed at pouring together the solutions containing 20.8 g of barium chloride and 18.0 g of sodium sulphate?

solution
BaCl2 + Na2SO4 ® BaSO4Ї + 2NaCl

Calculation of product amount carries on the source substance, which is taken in the deficit.
a)     Beforehand define which substance stand in the deficit.
Mark the amount of Na2SO4 – X
208 g (1 g-mol) BaCl2 react with 132 g (1 g-mol) Na2SO4
20.8 g with X g

X =20.8 * 132—————208= 13.2 g Na2SO4

We found, that on the reaction with 20.8 g BaCl2 will be spend 13.2 g Na2SO4, but we have 18.0 g. Thus, sodium sulphate in the reaction is taken in the excess and further calculation is necessary to carry on the BaCl2, which is taken in deficit.


b)     Define quantity of BaSO4 precipitate. 208 g (1 g-mol) BaCl2 form 233 g BaSO4.
20.8 g –– Y g.



233 * 20.8


Y
=
—————
=
23.3 g


208