Question

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

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.

The mass of
`NH_4Cl` produced is, 110.7 grams.

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 = 0.0821 L.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 = 56.5 L

Now we have to calculate the moles of
`NH_4Cl`

As, 1 mole of HCl react with 1 mole of
`NH_4Cl`

So, 2.07 mole of HCl react with 2.07 mole of
`NH_4Cl`

Now we have to calculate the mass of
`NH_4Cl`

`\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl* \text{ Molar mass of }NH_4Cl`

Molar mass of
`NH_4Cl` = 53.5 g/mole

`\text{ Mass of }NH_4Cl=(2.07moles)* (53.5g/mole)=110.7g`

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

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

The mass of
`NH_4Cl` produced is, 110.7 grams.