Answer question no. 9 and question no. 11.

Asked by tps.mjmdr | 10th Jul, 2018, 10:27: PM

Expert Answer:

 
To identify the limiting reagent, first write the chemical reaction,
 
space space space Li subscript 2 straight O subscript open parentheses straight s close parentheses end subscript space space space plus space space straight H subscript 2 straight O subscript open parentheses straight g close parentheses end subscript space space space space rightwards arrow with blank on top space space 2 LiOH subscript open parentheses straight s close parentheses end subscript space

29.88 space straight g space space space space space space space space 18 space straight g space space space space space space space space space space space space space space 47.95 space straight g

As space 29.88 space straight g space space space Li subscript 2 straight O space gives space space 47.95 space straight g space LiOH

space 65 space kg space equals space 65000 space straight g space of space space space Li subscript 2 straight O

Then comma space
space 65000 space straight g space of space space space Li subscript 2 straight O space will space give space space fraction numerator 47.95 cross times 65000 over denominator 29.88 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 104308.90 space straight g space of space space LiOH

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 104.3 space kg space space of space space LiOH

80 space kg space space equals space 80 comma 000 space straight g space of space straight H subscript 2 straight O

18 space straight g space straight H subscript 2 straight O space gives space space 47.95 space straight g space space LiOH

80000 space straight g space straight H subscript 2 straight O space will space give space space space fraction numerator 80000 cross times 47.95 over denominator 18 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 213111.1 space straight g

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 213.1 space Kg

As space the space space space Li subscript 2 straight O space is space forming space less space quantity space of space LiOH space then space water comma space space space Li subscript 2 straight O space is space straight a space limiting space reagent.

104.3 space kg space space of space space LiOH space is space formed space by space fraction numerator 104.3 cross times 18 over denominator 47.95 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 39.39 space kg

Excess space water space left space equals space 80 space minus space 39.39 space

space space space space space space space space space space space space space space space space space space space space space space space space space space equals 40.6 space kg space of space straight H subscript 2 straight O
 
Limiting reagent is Li2O
 
Excess of reactant remain is 40.6 kg.
 
 
Molality is preferred over molality because:
 
Molarity is the number of moles of solute present in one litre of solution.
 
Molality is the number of moles of solute present in 1 kg of solvent.
 
In the case of molarity, the change in temperature causes the change in volume of the solution but in molality, the temperature will not affect the weight of the solvent.
 
Therefore molality is preferred over molarity in expressing the concentration of a solution.

Answered by Varsha | 13th Jul, 2018, 12:49: PM