Question

What is the mass of solid NH4Cl formed when 75.5g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas remaining, measured at 14.0C and 752 mmHg? What gas is it? The formula is NH3 (g) + HCl gas -> NH4Cl solid

Answer 1

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:

`NH_3+HCl\rightarrow NH_4Cl`

First we have to calculate the moles of
`NH_3` and HCl.

`\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}`

Molar mass of
`NH_3` = 17 g/mole

`\text{Moles of }NH_3=(75.5g)/(17g/mole)=4.44mole`

and,

`\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}`

Molar mass of
`HCl` = 36.5 g/mole

`\text{Moles of }HCl=(75.5g)/(36.5g/mole)=2.07mole`

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of
`HCl` react with 1 mole of
`NH_3`

So, 2.07 mole of
`HCl` react with 2.07 mole of
`NH_3`

From this we conclude that,
`NH_3` is an excess reagent because the given moles are greater than the required moles and
`HCl` is a limiting reagent and it limits the formation of product.

The remaining moles of
`HCl` gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

`PV=nRT`

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 =
`0.0821L.atm/mol.K`

T = Temperature of gas =
`14.0^oC=273+14.0=287K`

Putting values in above equation, we get:

`0.989atm* V=2.37mole* (0.0821L.atm/mol.K)* 287K`

`V=56.5L`

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

Answer 2

Answer:The volume of the remaining gas that is ammonia is 23.85 L.

Step-by-step explanation:

`NH_3(g)+HCl(g)\rightarrow NH_4Cl(s)`

Moles of
`NH_3=\frac{\text{mass of} NH_3}{\text{Molar mass of}NH_3}=(75.5 g)/(17.03 g/mol)=4.43 mole`

Moles of HCl of gas =
`\frac{\text{mass of} HCl}{\text{Molar mass of}HCl}=(75.5 g)/(36.5 g/mol)=2.06 mol`

According to reaction 1 mole of HCl reacts with 1 mol of
`NH_3` then 2.06 moles of HCl will react with = 2.06 moles of
`NH_3`

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)

`V=(2.36 mol* 0.08206 L atm/K mol* 287 K)/(2.33 atm)=23.85 L`

The volume of the remaining gas that is ammonia is 23.85 L.