Critical Thinking
Category : NEET
Critical Thinking
In many situations, an excess of one or more substance is available for chemical reaction. Some of these excess substances will therefore be left over when the reaction is complete; the reaction stops immediately as soon as one of the reactant is totally consumed.
The substance that is totally consumed in a reaction is called limiting reagent because it determines or limits the amount of product. The other reactant present in excess are called as excess reagents.
Let us consider a chemical reaction which is initiated by passing a spark through a reaction vessel containing 10 mole of H2 and 7 mole of O2.
\[2\underset{{}}{\mathop{\,{{H}_{2}}\,}}\,(g)\,\,+\,\,\underset{{}}{\mathop{{{O}_{2}}\,}}\,(g)\,\xrightarrow{{}}\,\,2\,\,\underset{{}}{\mathop{{{H}_{2}}O}}\,\,(v)\]
Moles before reaction 10 7 0
Moles after reaction 0 2 10
The reaction stops only after consumption of 5 moles of O2 as no further amount of H2 is left to react with unreacted O2. Thus H2 is a limiting reagent in this reaction
Examples based on Stoichiometry and Limiting reagent
Example: 17 When a solution containing 4.77 gm. of NaCl is added to a solution of 5.77 gm. of AgNO3, the weight of precipitated AgCl is [IIT 1978]
(a) 11.70 gm. (b) 9.70 gm. (c) 4.86 gm. (d) 2.86 gm.
Solution: (c) \[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,AgN{{O}_{3}}\,\,\,\,\,\,\,\,+\,\,\,\,\,\,NaCl\,\,\xrightarrow{{}}\,\,AgCl\,\,\,\,\,\,\,+\,\,\,\,\,\,NaN{{O}_{3}}\]
Moles before mixing \[\frac{5.77}{108\,+\,14\,+\,48}\] \[\frac{4.77}{58.5}\] 0 0
= 0.0339 = 0.0815 (Here \[AgN{{O}_{3}}\]is limiting reactant, thus)
Moles after mixing 0 0.0815 ? 0.0339 0.0339 0.0339
= 0.0476
\[\therefore \] Moles of \[AgCl\] formed = 0.0339
\[\therefore \] Mass of \[AgCl\] formed = \[\text{Mol}\text{. mass}\,\,\times \,\,\text{No}\text{. of moles}\] = \[143.5\times 0.0339\] = \[4.864\,gm.\]
Example: 18 The volume of oxygen at STP required to completely burn 30 ml of acetylene at STP is [Orissa JEE 1997]
(a) 100 ml (b) 75 ml (c) 50 ml (d) 25 ml
Solution: (b) The balanced chemical equation for the reaction can be written as:
\[\underset{{}}{\mathop{{{C}_{2}}\,{{H}_{2}}}}\,\,\,+\,\,\,\,5/2\,\,{{O}_{2}}\,\xrightarrow{{}}\,2\,C{{O}_{2}}\,+\,{{H}_{2}}O\]
\[1\,Vol.\,\,\,\,\,\,\,\,\,5/2\,Vol.\]
\[1\,ml\,\,\,\,\,\,\,\,\,\,\,5/2\,ml\]
\[30\,ml\,\,\,\,\,\,30\,\times \,\,5/2\,=\,75\,ml\]
Hence, volume of the oxygen at STP required to burn 30 ml of acetylene at STP = 75 ml.
Example: 19 What is the volume (in litres) of oxygen at STP required for complete combustion of 32 g of\[C{{H}_{4}}\]
[EAMCET 2001]
(a) 44.8 (b) 89.6 (c) 22.4 (d) 179.2
Solution: (b) According to Avogadro's hypothesis, volume occupied by one mole of any gas at STP is 22.4 litres.
\[C{{H}_{4}}_{(g)}\,\,\,\,\,\ \ \,\,\,+\,\ \ \,\,\,\,\,\,\,\,2\,{{O}_{2}}_{(g)}\,\,\,\xrightarrow{{}}\,\,\,C{{O}_{2}}_{(g)}\,\,\,\,+\,\,\,\,\,2\,{{H}_{2}}{{O}_{\,(l)}}\]
1 mole 2 moles
2 moles 4 moles
\[2\,\times \,16\,gm.\,=\,32\,gm.\] \[4\,\times \,22.4\,\,\text{litres}\,=\,\text{89}\text{.6 litres}\]
Example: 20 A metal oxide has the formula \[{{Z}_{2}}{{O}_{3}}\]. It can be reduced by hydrogen to give free metal and water. 0.1596 g
of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is
[CBSE PMT 1989]
(a) 27.9 (b) 159.6 (c) 79.8 (d) 55.8
Solution: (d) Valency of metal in \[{{Z}_{2}}{{O}_{3}}=3\]
\[{{Z}_{2}}{{O}_{3}}+3{{H}_{2}}\xrightarrow{{}}2Z+3{{H}_{2}}O\]
\[0.1596\ gm\]of \[{{Z}_{2}}{{O}_{3}}\] react with \[{{H}_{2}}\] = \[6mg=0.006\ gm\]
\[1\ gm\] of \[{{H}_{2}}\] react with \[{{Z}_{2}}{{O}_{3}}\] \[=\frac{0.1596}{0.006}=26.6\ gm\]
Equivalent wt. of \[{{Z}_{2}}{{O}_{3}}=26.6=\] equivalent wt. of \[Z+\] equivalent wt. of \[O\]\[=E+8=26.6\] or \[E=18.6\]
Atomic weight of \[Z=18.6\times 3=55.8\]
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