How many grams of NaOH are needed to make 500 mL of a 2.5 M NaOH solution
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Answer:
Explanation:
First write the balance eqation of chemical reaction:
Remember writing any chemical reaction from its elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.
mass of oxygen given=5gram
mole of oxygen
mass of sulpher given=6gram
mole of sulpher
from the above balanced equaion;
2 mole of sulphur reacts with 3 mole Oxygen completely
1 mole of sulphur reacts with 1.5 mole Oxygen completely
0.19 mole of sulphur reacts with i.e. 0.285 mole Oxygen completely.
but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent
so product will depend on the limiting reagent
from the balance equation
3 mole of sulpher gives 2 mole
1 mole of sulpher will give 2/3 mole
0.16 of sulpher will give 0.12 mole
mass of =9.6gram this is theoreical production of
and
actual production of =7.9gram