
0.01 mole of iodoform ($CH{{I}_{3}}$) reacts with Ag powder to produce a gas whose volume at NTP is
(A) 224 ml
(B) 112 ml
(C) 336 ml
(D) 448 ml
Answer
576.3k+ views
Hint: By writing the reaction between iodoform and Ag powder, we will be able to get the stoichiometric ratio between reactants and products. From there we will be able to calculate the number of moles of gas and from that value, we can find the volume of gas.
Complete step by step solution:
-Let's start with the reaction between Ag powder and iodoform. When iodoform is allowed to react with ag powder acetylene gas is produced. The chemical reaction can be written as follows
$2CH{{I}_{3}}+6Ag\to {{C}_{2}}{{H}_{2}}+6AgI$
From the reaction, it is clear that, when iodoform is reacted with Ag powder, acetylene and silver iodide are formed.
-We are asked to find out about the volume of acetylene formed when 0.01 mole of iodoform reacts with Ag powder. So, let's look at the stoichiometry of the above reaction. When two moles of iodoform are reacted with six moles of Ag powder, one mole of acetylene and six moles of silver iodide are formed.
-Two moles of iodoform leads to one mole of acetylene. Hence the number of moles of acetylene produced by 0.01 moles of iodoform can be calculated as follows
$\dfrac{1}{2}\times 0.01=0.005moles$
That is 0.01 mole of iodoform gives 0.005 moles of acetylene.
-It given in the question the volume we need to calculate is at normal temperature pressure (NTP) conditions. As we know the volume of one mole of a gas is 22.4 litres. Hence the volume of 0.005 moles of acetylene can be calculated as follows
The volume of 1 mole of gas at NTP =22.4 litres
The volume of 0.005 moles of gas at NTP
$=22.4×0.005$
$= 0.112 litres$
$= 112 ml $
Therefore, the answer is option (B) 112 ml
Note: The answer can also be found by another numerical method in which the ideal gas equation is used. Since it’s given as an NTP condition we know about the number of moles, pressure and temperature and by substituting these known values into the equation $pV=nRT$ we will be able to find the volume of acetylene gas.
Complete step by step solution:
-Let's start with the reaction between Ag powder and iodoform. When iodoform is allowed to react with ag powder acetylene gas is produced. The chemical reaction can be written as follows
$2CH{{I}_{3}}+6Ag\to {{C}_{2}}{{H}_{2}}+6AgI$
From the reaction, it is clear that, when iodoform is reacted with Ag powder, acetylene and silver iodide are formed.
-We are asked to find out about the volume of acetylene formed when 0.01 mole of iodoform reacts with Ag powder. So, let's look at the stoichiometry of the above reaction. When two moles of iodoform are reacted with six moles of Ag powder, one mole of acetylene and six moles of silver iodide are formed.
-Two moles of iodoform leads to one mole of acetylene. Hence the number of moles of acetylene produced by 0.01 moles of iodoform can be calculated as follows
$\dfrac{1}{2}\times 0.01=0.005moles$
That is 0.01 mole of iodoform gives 0.005 moles of acetylene.
-It given in the question the volume we need to calculate is at normal temperature pressure (NTP) conditions. As we know the volume of one mole of a gas is 22.4 litres. Hence the volume of 0.005 moles of acetylene can be calculated as follows
The volume of 1 mole of gas at NTP =22.4 litres
The volume of 0.005 moles of gas at NTP
$=22.4×0.005$
$= 0.112 litres$
$= 112 ml $
Therefore, the answer is option (B) 112 ml
Note: The answer can also be found by another numerical method in which the ideal gas equation is used. Since it’s given as an NTP condition we know about the number of moles, pressure and temperature and by substituting these known values into the equation $pV=nRT$ we will be able to find the volume of acetylene gas.
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