Question

For given reaction,
$CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O + C{O_2}$
The volume of $C{O_2}$ gas formed when $2.5g$ calcium carbonate is dissolved in excess hydrochloric acid at ${0^ \circ }C$ and $1atm$ pressure is:
[1 mole of any gas at ${0^ \circ }C$ and $1atm$ pressure occupies $22.414L$ volume]
A) $1.12L$
B) $56.0L$
C) $0.28L$
D) $0.56L$

Answer 1

Hint: We perceive the volume involved by one mole of substance at a given temperature and weight is called molar volume. It is routinely valuable to the gases where the idea of the gas doesn't influence the volume. The most widely recognized representation is that the molar volume of gas at standard temperature-pressure condition is equivalent to $22.4L$ for one mole of an ideal gas at temperature equivalent to $273K$ and pressure equivalent to $1\,atm$.

Complete step by step answer:
The reaction is $CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O + C{O_2}$
We are on familiar terms with that, At STP, 1 mole of a gas occupies \[22.4L\] of volume.
Thus volume of carbon dioxide formed from $2.5gCaC{O_3}$ can be calculated as,
$Volume = \dfrac{{2.5 \times 22.4}}{{100}} = 0.56L$
Thus the volume of carbon dioxide formed from $2.5gCaC{O_3}$ is $0.56L$.
Therefore, the option (D) is correct.

Additional information:
We realize that, Vapor density is the proportion of the mass of a volume of a gas, to the mass of an equivalent volume of hydrogen, estimated under the standard states of temperature and pressure.
${\text{Vapour density}} = \dfrac{{{\text{Mass of n molecules of gas}}}}{{{\text{Mass of n molecule of hydrogen}}}}$
First, we have to calculate the molar mass of a gas using the relation.
\[\text{Molar Mass} = 2 \times \text{vapor density}\]
If the given vapor density of the gas is \[11.2\].
Thus, the molar mass of the gas is calculated as,
\[\text{Molar Mass} = 2 \times 11.2 = 22.4gm/mole\]
Next, we have to calculate the number of moles of the gas.
Let us assume that the given amount of the gas is \[{\text{24}}gm\].
Therefore the number of moles of the gas is given by,
\[\text{Number Of Moles} = \dfrac{{2.4}}{{22.4}}mole = 0.1071moles\]
We are on familiar terms with that, At STP, 1 mole of a gas occupies \[22.4L\] of volume. Here, we have about $0.1071moles$ of the gas. Hence, the volume of gas filled at STP is,
\[\text{Volume Occupied} = 0.1071 \times 22.4{\text{L}} = \;2.4L\]
Thus, $24g$ of a gas, with a vapor density of \[11.2\], will occupy \[2.4L\] of volume at STP.

Note:
Now we can see the difference between atmosphere and NTP:
Standard temperature and pressure conditions are thought of as STP. The quality temperature value is \[0^\circ C\] and the standard pressure value is \[100kPa\] or \[1bar\] Normal Temperature and Pressure is known as NTP the worth of pressure at NTP is \[101.325kPa\] and the temperature at NTP is \[20^\circ C\].