Question

The reaction is ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}\xrightarrow{{}}C{r_2}{O_3} + 4{H_2}O + {N_2}$
Calculate the volume of nitrogen at STP, evolved when $63g$ of ammonium dichromate is heated.
A) $5.6litre$
B) $11.2litre$
C) $22.4litre$
D) $18.4litre$

Answer 1

Hint: We recognize the volume occupied by one mole of substance at a given temperature and pressure is called molar volume. It is regularly useful to the gases where the nature of the gas does not affect the volume. The most common illustration is that the molar volume of gas at standard temperature-pressure condition is equal to $22.4L$ for one mole of an ideal gas at temperature equal to $273K$ and pressure equal to$1atm$.

Formula used:

Complete step by step answer:
We can write the given reaction as, ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}\xrightarrow{{}}C{r_2}{O_3} + 4{H_2}O + {N_2}$
We know that the molecular weight of ammonium dichromate is$252\,g$.
Now, $252g$ of ammonium dichromate emits $22.4L{N_2}$.
The amount of aluminium dichromate evolved by $63g = \dfrac{{22.4 \times 63}}{{252}} = 5.6L$
The amount of aluminium dichromate evolved by $63g$ is $5.6L$.

So, the correct answer is Option A.

Additional Information:
We recognize that, vapour density is the quantitative relation of the mass of a volume of a gas, to the mass of an associate equal volume of hydrogen, measured below the quality conditions of temperature and pressure.
First, we get the molar weight of a gas.
The molar weight of a gas, by known values the vapor density is calculated by the formula:
${\text{Molar mass}} = 2 \times {\text{Vapour density}}$
The vapor density of the gas is\[11.2\].
Thus, the molar mass of the gas is calculated as,
Molar mass\[ = 2 \times 11.2 = 22.4gm/mole\]
The given amount of the gas is\[{\text{24}}gm\].
Therefore the amount of moles of the gas is calculated as,
No. of moles \[ = {\text{2}}{\text{.4}}/22.4mole = 0.1071moles\]
We recognize that, At STP, 1 mole of a gas occupies \[22.4{\text{L}}\] of volume. Here, we have about $0.1071moles$ of the gas. Therefore, the volume occupied by the gas at STP is,
Volume occupied \[ = 0.1071 \times 22.4{\text{L}} = 2.4L\]
Thus, the $24g$ of a gas, with a vapor density of \[11.2\], will occupy \[2.4L\] of volume at STP

Note: Now we discuss about the difference between atmosphere and NTP as,
We must remember that the Standard temperature and pressure condition is 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 \[20^\circ C\].