Answer :
Answer : The total volume of gas produced are, 11.5 L
Explanation :
First we have to calculate the moles of ammonium carbonate.
[tex]\text{Moles of }(NH_4)_2CO_3=\frac{\text{Mass of }(NH_4)_2CO_3}{\text{Molar mass of }(NH_4)_2CO_3}[/tex]
Molar mass of [tex](NH_4)_2CO_3[/tex] = 96.094 g/mol
[tex]\text{Moles of }(NH_4)_2CO_3=\frac{11.9g}{96.094g/mol}[/tex]
[tex]\text{Moles of }(NH_4)_2CO_3=0.124mol[/tex]
Now we have to calculate the moles of total gas.
The given balanced chemical reaction is:
[tex](NH_4)_2CO_3(s)\rightarrow 2NH_3(g)+CO_2(g)+H_2O(g)[/tex]
From the balanced chemical reaction we conclude that,
As, 1 mole of [tex](NH_4)_2CO_3[/tex] react to give 4 mole of gas
So, 0.124 mole of [tex](NH_4)_2CO_3[/tex] react to give [tex]0.124\times 4=0.496[/tex] mole of gas
Now we have to calculate the volume of gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of gas = 1.05 atm
V = Volume of gas = ?
n = number of moles = 0.496 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of gas = [tex]24.0^oC=273+24.0=297.0K[/tex]
Putting values in above equation, we get:
[tex]1.05atm\times V=0.496mole\times (0.0821L.atm/mol.K)\times 297.0K[/tex]
[tex]V=11.5L[/tex]
Thus, the total volume of gas produced are, 11.5 L
The total volume of all the gases produced in the combustion of 11.9g ammonium carbonate has been 12.02 L.
From the balanced chemical equation, 1 moles of ammonium carbonate produced 2 moles ammonia, 1 mole carbon dioxide, and 1 mole water.
The available ammonium carbonate has been 11.9 grams. The moles of ammonium carbonate has been:
Moles = [tex]\rm \dfrac{Weight}{Molecular\;weight}[/tex]
Moles of ammonium carbonate = [tex]\rm \dfrac{11.9\;g}{96.09\;g/mol}[/tex]
Moles of ammonium carbonate = 0.123 mol.
The moles of gases produced by 0.123 mol of ammonium carbonate has been:
- 1 mole ammonium carbonate = 2 moles ammonia
0.123 moles ammonium carbonate = 0.247 moles ammonia
- 1 mole ammonium carbonate = 1 mole carbon dioxide
0.123 moles ammonium carbonate = 0.123 moles carbon dioxide
- 1 moles ammonium carbonate = 1 moles water
0.123 moles ammonium carbonate = 0.123 moles water.
The total moles of gases produced:
= Moles of ammonia + moles of carbon dioxide + moles of water
= 0.247 + 0.123 + 0.123
= 0.493 moles.
According to the ideal gas equation,
PV = nRT
P = pressure = 1 atm.
V = volume
n = moles of gas = 0.493 mol.
R = constant = 0.08206 L.atm/mol. K
T = temperature = 297.15 K.
Substituting the values:
1 × Volume = 0.493 × 0.08206 × 297.15
Volume = 12.02 L.
The total volume of all the gases produced in the combustion of 11.9g ammonium carbonate has been 12.02 L.
For more information about the volume of gas, refer to the link:
https://brainly.com/question/1218415