Question

The volume of gas at NTP produced by $100\,g$ of $Ca{C_2}$ with water is.
A)$70\,L$
B)$35\,L$
C)$17.5\,L$
D)$22.4\,L$

Answer 1

We know that,
The number of moles can be calculated using the formula,
$Moles = \dfrac{{Mass\left( g \right)}}{{Molecular\,Mass\left( {g/mol} \right)}}$

Complete step by step answer:
First, write the balanced chemical equation of the reaction,
$Ca{C_2} + 2{H_2}O\xrightarrow{{}}Ca{\left( {OH} \right)_2} + {C_2}{H_2}$
Given: The mass of calcium carbide is$100\,g$
We know the molecular weight of calcium carbide is $64\,g/mol.$
The number of moles of calcium carbide is calculated using the formula,
$Moles = \dfrac{{Mass\left( g \right)}}{{Molecular\,Mass\left( {g/mol} \right)}}$
$Moles = \dfrac{{100\,g}}{{64\,g/mol}} = 1.56\,mol$
The number of moles of calcium carbide is$1.56\,mole$.
Molar volume at NTP can be used to convert from moles to gas volume by using the conversion factor$1mol = 22.4L$.
One mole of calcium carbide gives $22.4\,L$ of acetylene at NTP.
Thus, $1.56\,mole$ of calcium carbide gives,
$1.56\,mole \times \dfrac{{22.4\,L}}{{1\,mole}} = 35\,L$
The volume of gas at NTP produced by $100\,g$ of $Ca{C_2}$ with water is $35\,L$.

So, the correct answer is Option B.

Additional Information:
Molar volume:
The volume occupied by one mole of substance at a given temperature and pressure is called molar volume. It is usually applied to the gases where the nature of the gas does not affect the volume. The most general example is that the molar volume of gas at standard temperature-pressure condition is equal to $22.4\,L$ for one mole of an ideal gas at temperature equal to $273\,K$ and pressure equal to$1\,atm$.

Note:
Now we can see the difference between STP and NTP. Standard temperature and pressure condition is known as STP. The standard temperature value is $0^\circ C$ and the standard pressure value is $100{\text{ }}kPa$ or $1{\text{ }}bar.$ Normal Temperature and Pressure is known as NTP the value of pressure at NTP is $101.325{\text{ }}kPa$ and the temperature at NTP is $20^\circ C$.