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Stoichiometric Calculations

CO3 3+6 NaNO ...

Question

For the reaction $2 F e (N O_{3})_{3} + 3 N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + 6 N a N O_{3}$ , initially 2.5 mol of $F e (N O_{3})_{3}$ and 3.6 mol of $N a_{2} C O_{3}$ is taken. If 6.3 mol of $N a N O_{3}$ is obtained then, % yield of given reaction is:

50 %

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84 %

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87.5 %

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100 %

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Solution

The correct option is C $87.5 %$
$2 F e (N O_{3})_{3} 2.5 m o l e + 3 N a_{2} C O_{3} 3.6 m o l e \to F e_{2} (C O_{3})_{3} + 6 N a N O_{3} 6.3 m o l e$
To find limiting reagent
for $F e (N O_{3})_{3} = \frac{2.5}{2} = 1.25$
for $N a_{2} C O_{3} = \frac{3.6}{3} = 1.2$
Here the limiting reagent is $N a_{2} C O_{3}$ , so $N a N O_{3}$ formed (theoretically) will be $\frac{3.6 m o l o f N a_{2} C O_{3} \times 6 m o l o f N a N O_{3}}{3 m o l o f N a_{2} C O_{3}} = 7.2 m o l o f N a N O_{3}$
$P e r c e n t a g e y i e l d = \frac{A c t u a l y i e l d}{t h e o r e t i c a l y i e l d} \times 100 = \frac{6.3}{7.2} \times 100 = 87.5 %$

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Stoichiometric Calculations

Standard XII Chemistry

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For the reaction 2Fe(NO3)3+3Na2CO3→Fe2(CO3)3+6NaNO3, initially 2.5 mol of Fe(NO3)3 and 3.6 mol of Na2CO3 is taken. If 6.3 mol of NaNO3 is obtained then, % yield of given reaction is:

For the reaction $2 F e (N O_{3})_{3} + 3 N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + 6 N a N O_{3}$ , initially 2.5 mol of $F e (N O_{3})_{3}$ and 3.6 mol of $N a_{2} C O_{3}$ is taken. If 6.3 mol of $N a N O_{3}$ is obtained then, % yield of given reaction is: