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Stoichiometric Calculations

Consider the ...

Question

Consider the following chemical reaction:
$F e S_{2} + N a_{2} O_{2} + H_{2} S O_{4} \to N a_{2} F e O_{4} + N a_{2} S O_{4} + H_{2} O$
Find the correct option(s) regarding the above reaction.

To form 24 mol of

N a_{2} S O_{4}

27 moles of

N a_{2} O_{2}

are required

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n-factor of

F e S_{2}

is 18

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On complete reaction of 36 equivalents of

F e S_{2}

, moles of

N a_{2} S O_{4}

formed is 16

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Oxidation state of Fe in the reactant compound is +4

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Solution

The correct options are
A To form 24 mol of $N a_{2} S O_{4}$ 27 moles of $N a_{2} O_{2}$ are required
B n-factor of $F e S_{2}$ is 18
C On complete reaction of 36 equivalents of $F e S_{2}$ , moles of $N a_{2} S O_{4}$ formed is 16
Balanced reaction is
$F e S_{2} + 9 N a_{2} O_{2} + 6 H_{2} S O_{4} \to N a_{2} F e O_{4} + 8 N a_{2} S O_{4} + 6 H_{2} O$
a) From the balanced chemical equation, for 8 mol of $N a_{2} S O_{4}$ , 9 mol of of $N a_{2} O_{2}$ is required. Thus to form 24 mol of $N a_{2} S O_{4}$ number of mol of $N a_{2} O_{2}$ required is 27 mol.
b) The oxidation state of $F e$ is changing from +2 to +3 and
oxidation state of $S$ is changing from -1 to +6. Thus the n factor for $F e S_{2}$ is $1 \times 4 + 2 \times 7 = 18$ .
c) For 16 mol of $N a_{2} S O_{4}$ , 2 mol of $F e S_{2}$ is required.
So equivalent of $F e S_{2}$ = $m o l \times (n - f a c t o r) = 36 e q u v$
d) Oxidation state of Fe in $F e S_{2}$ is +2

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