Sr(NO 3 ) 2 (aq) + 2 KIO 3 (aq) + H 2 O(l) 2 KNO 3 (aq) + Sr(IO 3 ) 2 ·H 2 O(s)

Question

Sr(NO3)2 (aq) + 2 KIO3 (aq) + H2O(l) 2 KNO3(aq) + Sr(IO3)2·H2O(s)

At the end of this experiment the weight of strontium iodate monohydrate is taken.

Sr(NO3)2 = 211.6298 g/mol
KIO3 = 214 g/mol
Sr (IO3)2.H2O = 455.44 g/mol
KNO3 = 101.1032 g/mol
H2O = 18.015 g/mol

QUESTION: Determine the LIMITING REACTANT and the THEORETICAL YIELD of strontium iodate monohydrate. You are given the following for TWO trials: (Calculate each trial)

Trial 1

Trial 2

KIO3

Sr(NO3)2

KIO3

Sr(NO3)2

Mass (g)

1.500

1.500

1.800

1.800

Volume DI water (mL)

50

25

50

25

Trial 1

Trial 2

KIO3

Sr(NO3)2

KIO3

Sr(NO3)2

Mass (g)

1.500

1.500

1.800

1.800

Volume DI water (mL)

50

25

50

25

Explanation / Answer

Sr(NO3)2 (aq) + 2KIO3(aq) + H2O(l) --------> 2KNO3(aq) + Sr(IO3)2.H2O (s)

Stoichiometry masses

Sr(NO3)2 = 211.6298g/mol × 1 = 211.6298g

KIO3 = 214g/mol × 2 = 428g/mol

H2O(l) = 18g/mol× 1 =18g

Sr(IO3)2.H2O = 455.44g/mol × 1 = 455.44g/mol

Experimental masses

Trial 1

KIO3 = 1.500g

Sr(NO3)2 = 1.500g

H2O = 75g

Stoichiometrically 211.6298g of Sr(NO3)2 require 428g of KIO3 and 18g of H2O So 1.500g of Sr(NO3)2 require 3.0335g of KIO3 and 0.1276g of H2O but available masses of KIO3 is 1.50 and available mass of H2O is 75g

So , Sr(NO3)2 is Limiting

211.6298g of Sr(NO3)2 give 455.44g of Sr(IO3)2.H2O

So , 1.5g of Sr(NO3)2 give (455.44g/211.6298g)×1.5g = 3.2281g

Therefore,

Theoretical Yield = 3.2281g

Trial 2

Similar to the above calculation

Limiting reagent is Sr(NO3)2

Theoretical Yield = 3.8737g

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.