Question

Solid ammonium chloride, NH, CI, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCI. NH3(g) + HCl(g) -NH,CH(s) A 4.93 g sample of NH3 gas and a 4.93 g sample of HCl gas are mixed in a 1.50 L flask at 25°C Identify the limiting reagent. O NH3 Она O NHACI How many grams of NHACI will be formed by this reaction? mass: What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NHCl produced by the reaction

Answer 1

The balanced equation is

NH₃ + HCl --------> NH₄Cl

Number of moles of NH₃ =mass/mol.wt. = 4.93 g/(17.03 g/mol) = 0.289489 mol

Number of moles of HCl =mass/mol.wt. = 4.93 g/(36.46 g/mol) = 0.135217 mol

According to balanced equation,

1 mole of HCl reacts with 1 mole of NH₃

So, 0.135217 mol of HCl would react with 0.135217 mol of NH₃

But the number of moles of NH₃ present is more than required.

That is NH₃ is the Excess reagent and HCl is the Limiting reagent.

According to balanced equation,

1 mole of HCl produces1 mole of NH₄Cl

So, 0.135217 mol of HCl would produce 0.135217 mol of NH₄Cl

Mass of NH₄Cl = moles * mol.wt. = 0.135217mol * 53.49 g/mol = 7.23274 g

Mass of NH₄Cl = 7.23 g

(c)

According to balanced equation,

1 mole of HCl reacts with 1 mole of NH₃

So, 0.135217 mol of HCl would react with 0.135217 mol of NH₃

But the number of moles of NH₃ present is more than required.

That is NH₃ is the Excess reagent

Number of moles of NH₃ present in excess = 0.289489 – 0.135217 = 0.154272 mol

No other gases is present other than NH₃ in the flask

Number of moles = 0.154272 mol

Volume = 1.50 L

Temperature = 25 °C = 25 + 273.15 = 298.15 K

We know that

PV = nRT

P = nRT/V = (0.154272 mol * 0.0821 L.atm.mol-1.K-1 * 298.15 K)/(1.50 L

P = 2.517533 atm

Pressure in the flask = 2.52 atm

Answer:

Limiting reagent = HCl

mass of NH₄Cl = 7.23 g

Pressure in the flask = 2.52 atm