Question

The volume of the gas produced by 100 g of $\text {Ca}{{\text{C}} _ {2}} $with water is:
$\text{Ca}{{\text{C}}_{2}}\text{+}{{\text{H}}_{2}}\text{O}\to \text{ Ca(OH}{{\text{)}}_{2}}\text{+}{{\text{C}}_{2}}{{\text{H}}_{2}}$
(A) 70L
(B) 35L
(C) 17.5L
(D) 22.4L

Answer 1

Hint: It obeys Avogadro’s law in which 1mole of gas occupies volume of 22.4 litres at NTP and by finding the no of moles of $\text{Ca}{{\text{C}}_{2}}$using the given mass and the molar mass of $\text{Ca}{{\text{C}}_{2}}$,we can easily calculate the volume of gas produced.


Complete step by step answer:
This numerical is based on the Avogadro’s law which states that the equal volumes of the gases at the same temperature and pressure contains equal number of molecules i.e.$\text {6} \text {.023} $×${{10} ^ {23}} $Avogadro’s particles and 1 mole of every substance occupies 22.4 litres at NTP (i.e. at normal temperature pressure).
Now, consider the above given equation:
$\text{Ca}{{\text{C}}_{2}}\text{+}{{\text{H}}_{2}}\text{O}\to \text{ Ca(OH}{{\text{)}}_{2}}\text{+}{{\text{C}}_{2}}{{\text{H}}_{2}}$ ------------(A)
1mole 22.4L
i.e. 1mole of $\text {Ca}{{\text{C}} _ {2}} $gives 22.4L of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $at NTP.
In the statement we have been given the mass of $\text {Ca}{{\text{C}} _ {2}} $=100g.
Now, first we have to convert it into moles and we know that:
  Number of moles of $\text {Ca}{{\text{C}} _ {2}} $=$\dfrac {\text {given mass}} {\text {molar mass}} $ --------(i)
Given mass of $\text {Ca}{{\text{C}} _ {2}} $=100g
Molar mass of $\text {Ca}{{\text{C}} _ {2}} $= 40×24
 = 64g
Put all these values in equation (i), we get:
 Number of moles of $\text {Ca}{{\text{C}} _ {2}} $=$\dfrac {100}{64} $
 =1.56 moles of $\text {Ca}{{\text{C}} _ {2}} $
Now, considering the above equation (A);
$\text{Ca}{{\text{C}}_{2}}\text{+}{{\text{H}}_{2}}\text{O}\to \text{ Ca(OH}{{\text{)}}_{2}}\text{+}{{\text{C}}_{2}}{{\text{H}}_{2}}$
1mole of $\text {Ca}{{\text{C}} _ {2}} $gives = 22.4L of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $
1.56 moles of $\text {Ca}{{\text{C}} _ {2}} $gives= 22.4×1.56 L of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $
=35L of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $at NTP
Therefore, the volume of the gas produced by 100 g of $\text {Ca}{{\text{C}} _ {2}} $with water is 35L of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $at NTP.


So, option (b) is correct.



Note: The given mass of any substance cannot be taken in place of the number of moles of that compound and is generally calculated by dividing the given mass with the molecular mass of that very compound and one mole of gas always occupies 22.4 l of volume at STP and NTP.