Respuesta :
Answer:The volume of the remaining gas that is ammonia is 23.85 L.
Explanation:
[tex]NH_3(g)+HCl(g)\rightarrow NH_4Cl(s)[/tex]
Moles of [tex]NH_3=\frac{\text{mass of} NH_3}{\text{Molar mass of}NH_3}=\frac{75.5 g}{17.03 g/mol}=4.43 mole[/tex]
Moles of HCl of gas = [tex]\frac{\text{mass of} HCl}{\text{Molar mass of}HCl}=\frac{75.5 g}{36.5 g/mol}=2.06 mol[/tex]
According to reaction 1 mole of HCl reacts with 1 mol of [tex]NH_3[/tex] then 2.06 moles of HCl will react with = 2.06 moles of [tex]NH_3[/tex]
Moles left of ammonia left = 4.43 - 2.06 = 2.36 moles
Volume of the gas will be given by Ideal gas equation: PV=nRT
Pressure = 752 mmHg = 752 × 0.0031 atm = 2.33 atm
R = 0.08026 L atm/K mol
V = ? , n = number of moles of ammonia
Temperature = 14 °C = 14 + 273 K = 287 K(0°C = 273K)
[tex]V=\frac{2.36 mol\times 0.08206 L atm/K mol\times 287 K}{2.33 atm}=23.85 L[/tex]
The volume of the remaining gas that is ammonia is 23.85 L.
Answer : The volume of the gas remaining is 56.5 liters.
The gas is hydrochloric acid and the formula of the gas is HCl.
Explanation :
The balanced chemical reaction will be:
[tex]NH_3+HCl\rightarrow NH_4Cl[/tex]
First we have to calculate the moles of [tex]NH_3[/tex] and HCl.
[tex]\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}[/tex]
Molar mass of [tex]NH_3[/tex] = 17 g/mole
[tex]\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole[/tex]
and,
[tex]\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}[/tex]
Molar mass of [tex]HCl[/tex] = 36.5 g/mole
[tex]\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole[/tex]
Now we have to calculate the limiting and excess reagent.
From the balanced reaction we conclude that
As, 1 mole of [tex]HCl[/tex] react with 1 mole of [tex]NH_3[/tex]
So, 2.07 mole of [tex]HCl[/tex] react with 2.07 mole of [tex]NH_3[/tex]
From this we conclude that, [tex]NH_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]HCl[/tex] is a limiting reagent and it limits the formation of product.
The remaining moles of [tex]HCl[/tex] gas = 4.44 - 2.07 = 2.37 moles
Now we have to calculate the volume of the gas remaining.
Using ideal gas equation :
[tex]PV=nRT[/tex]
where,
P = Pressure of gas = 752 mmHg = 0.989 atm (1 atm = 760 mmHg)
V = Volume of gas = ?
n = number of moles of gas = 2.37 moles
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of gas = [tex]14.0^oC=273+14.0=287K[/tex]
Putting values in above equation, we get:
[tex]0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K[/tex]
[tex]V=56.5L[/tex]
Thus, the volume of the gas remaining is 56.5 liters.
The gas is hydrochloric acid and the formula of the gas is HCl.
