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$CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O + C{O_2}$

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) $0.56L$

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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 $0.56L$.

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.

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 \[20^\circ C\].