You visited us 3 times! Enjoying our articles? Unlock Full Access!

Percentage Composition

For the react...

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 if $3$ moles of $F e (N O_{3})_{3}$ and $3.6$ moles of $N a_{2} C O_{3}$ is taken and $6.3$ moles of $N a N O_{3}$ is obtained, then the $%$ yield of given reaction is:

50 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

84 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

87.5 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $87.5 %$
$\begin{matrix} 2 F e (N O_{3})_{3} + & 3 N a_{2} C O_{3} & \to F e_{2} (C O_{3})_{3} & + & 6 N a N O_{3} Mole & 3 & 3.6 Mole/Stoichiometric coefficient & 1.5 & 1.2 \end{matrix}$
Limiting reagent is $N a_{2} C O_{3}$ .
So, moles of $N a N O_{3}$ should be formed $= 3.6 \times 2 = 7.2$
$% yield = \frac{6.3}{7.2} \times 100 = 87.5 %$ .

Suggest Corrections

Similar questions

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}

F e (N O_{3})_{2}

N a_{2} C O_{3}

N a N O_{3}

F e (N O_{3})_{3} + N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + N a N O_{3}

F e (N O_{3})_{2}

N a_{2} C O_{3}

N a N O_{3}

%

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}

F e (N O_{3})_{3}

N a_{2} C O_{3}

N a N O_{3}

2 X + 3 Y + 4 Z \to 5 W

1

3

4

1.25

7 A + 13 B + 15 C \to 17 P

Join BYJU'S Learning Program