MATH QUESTION HELP

Question

Solid ammonium chloride, NH4Cl,NH4Cl, is formed by the reaction of gaseous ammonia, NH3,NH3, and hydrogen chloride, HCl.HCl.

NH3(g)+HCl(g)⟶NH4Cl(s) NH3⁢(g)+HCl(g)⟶NH4Cl(s) 

A 4.59 g4.59 g sample of NH3NH3 gas and a 4.59 g4.59 g sample of HClHCl gas are mixed in a 1.00 L1.00 L flask at 25 ∘C.25⁢ ∘C.

Identify the limiting reagent.

HClHCl

NH3NH3

NH4ClNH4Cl

How many grams of NH4ClNH4Cl will be formed by this reaction?

mass:

g

What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH4ClNH4Cl produced by the reaction.

𝑃=�=

Answer 1

NH3(g) + HCl(g) ==> NH4Cl(s) .. Balanced equation

molar mass NH3 = 17 g / mol

molar mass HCl = 36.5 g / mol

molar mass NH4Cl = 53.5 g / mol

The limiting reactant can be found by dividing the moles of each reactant by the corresponding coefficient in the balanced equation and which ever value is less represents the limiting reactant.

For NH3: 4.59 g x 1 mol NH3/17 g = 0.27 mols NH3

For HCl: 4.59 g x 1 mol NHCl/36.5 g = 0126 mols HCl

Since both are divided by 1, the limiting reactant is HCl

Mass NH4Cl formed = 0.126 mols HCl x 1 mol NH4Cl / mol HCl x 53.5 g NH4Cl = 6.73 g NH4Cl

To find the pressure remaining at the end of the reaction, we must find mols of NH3(g) left over and then apply the ideal gas law. The pressure will be due only to NH3 since all the HCl(g) will have been used in the reaction.

moles NH3 used up = 0.126 mol HCl x 1 mol NH3 / mol HCl = 0.126 mols NH3 used

moles NH3 left over = 0.27 mols - 0. 126 mols = 0.144 mols NH3(g)

PV = nRT

P = nRT / V = (0.144)(0.0821)(298) / 1.00

P = 3.52 atm