Question

For the reaction, 2Fe(NO3)3 + 3Na2CO3 → Fe2(CO3)3 + 6NaNO3
Initially 2.5 mol of Fe(NO3)2 and 3.6 moles of Na,Co, are taken. If 6.3 mol of NaNo, is obtained then % yield
of given reaction is:
A) 50
B) 84
C) 87.5
D) 100​

Answer 1

According to the Reaction

$\boxed{\bold{\sf{2Fe(NO_3)_3 + 3Na_2CO_3 \rightarrow Fe_2(CO3)_3 + 6NaNO_3}}}$

Now,

Amount of Ferric Nitrate =2.5 Mole

Stoichiometric coefficient of Ferric Nitrate =2

Mole ratio will be = 2.5/2 = 1.25

Amount of sodium carbonate =3.6 moles

Stoichiometric coefficient = 3

Mole ratio = 3.6/3 = 1.2

Hence,

Sodium carbonate is Limiting Reagent in the reaction.

Now,

3 moles of sodium carbonate produces 6 moles of sodium Nitrate.

Thus,

1 mole of sodium carbonate will produce 2 moles of sodium nitrate

Therefore :-

3.6 mole of sodium carbonate produces 7.2 mole of Sodium Nitrate

The given yield of sodium Nitrate is 6.3 mole.

Thus,

Actual Yield = 6.3 moles

Theoretical Yield = 7.2 moles

So,

Percentage Yield will be:-

$= \frac{actual \: yield}{theoretical \: yield} \times 100\% \\ \\ \\ = \frac{6.3}{7.2} \times 100\% \\ \\ \\ = 87.5\%$

Therefore :-

Option (C) is correct!

Answer 2

$\mathfrak{\underline{\underline{Answer :-}}}$

$<b>The Percentage yield of given reaction is 87.5 %</b>$

$\mathfrak{\underline{\underline{Explanation :-}}}$

$\sf {Given \:reaction :}\\\\\\</p><p></p><p></p><p></p><p>\boxed {\sf {2 Fe(NO_3)_3 + 3 Na_2CO_3 \longrightarrow Fe_2 (CO)_3 + 6NaNO_3}}$

$\sf {2 \: mole \: of \: Fe(NO_3)_3 \: reacts \: with \: 3 \: moles \: of \: Na_2CO_3}\\\\\\ \sf {Now, we \: need \: to \: find \: how \: much \: mole \: of \: Na_2CO_3 \: will \: react} \\ \sf {with \: 2.5 \: mole \: of \: Fe(NO_3)_3} \\\\\\ \sf {2 \longrightarrow 3}\\\\ \sf {2.5 \longrightarrow x}\\\\\\ \textsf{By cross multiplication, we get :}\\\\\\</p><p> \sf {\implies 2x = 2.5 \times 3} \\\\\\ \sf {\implies x = 3.75} \\\\\\ \sf {\therefore 2.5 \: mole \: of \: Fe(NO_3)_3 \: will \: react \: with \: 3.75 \:mole \: of \: Na_2CO_3} \\\\\\ \sf {But \: we \: are \: given \: with \: only \: 3.6 \: moles \: of \: Na_2CO_3. So \: some}\\ \sf {amount \: of \: Fe(NO_3)_3 \: will \: be \: unreactive} \\\\\\ \sf {From \: the \: equation, 3 \: moles \: of \: Na_2CO_3 \: will \: produce \: 6 \: moles } \\ \sf{ of \: NaNO_3 } \\\\\\ \sf {Now, we \: need \: to \: find \: how \: much \: moles \: of \: NaNO_3} \\ \sf{will \: be \: produced \: from \: 3.6 \: moles \: of \: Na_2CO_3}\\\\\\ \sf {3 \longrightarrow 6}\\\\ \sf {3.6 \longrightarrow y}\\\\\\ \textsf{By cross multiplication, we get :}\\\\\\ \sf {\implies 3y = 3.6 \times 6} \\\\\\ \sf {\implies y = 7.2 } \\ \\ \\ \sf {\therefore 7.2 \: mole \: of \: NaNO_3 \: will \: be\: produced\: from \: 3.6 \:mole\:of\: Na_2CO_3}\\\\\\ \sf{\therefore Percentage\: Yield = \dfrac {Experimental\:yield}{Calculated \:yield} \times 100}\\\\\\\sf{\implies Percentage\: Yield = \dfrac {6.3}{7.2} \times 100}\\\\\\\sf{\implies Percentage\: Yield = 87.5 \%}$