Question

Ammonia and gaseous hydrogen chloride combine to form ammonium chloride. NH3+HCL->NH4CL

if 4.21 L of NH3(g) at 27 C and 1.02 atm is combined with 5.35 L of HCL(g) at 26 C and 0.998 atm what mass of NH4Cl(s) will be produced? which gas is the limiting reactant? which gas is present in excess?

Answer 1

To calculate the mass of ammonium chloride produced and determine the limiting reactant, the ideal gas law is used to find the moles of ammonia and hydrogen chloride. The reactant present in the lesser molar amount is the limiting reactant. Finally, convert the moles of the limiting reactant to grams using the molar mass of ammonium chloride.

Step-by-step explanation:

Calculating the Mass of Ammonium Chloride and Identifying the Limiting Reactant

The reaction between ammonia (NH3) and hydrogen chloride (HCl) to form ammonium chloride (NH4Cl) involves combining these gases in a one-to-one molar ratio according to the balanced chemical equation:
NH3(g) + HCl(g) → NH4Cl(s).

First, we need to use the ideal gas law to determine the moles of each gas. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvin.

For NH3: (1.02 atm)(4.21 L) = n(0.0821 L·atm/(mol·K))(300 K)
For HCl: (0.998 atm)(5.35 L) = n(0.0821 L·atm/(mol·K))(299 K)

After calculating the moles of each gas, compare the mole ratio to the balanced equation to identify the limiting reactant and the reactant in excess. The limiting reactant will determine the maximum amount of NH4Cl that can be produced. Then, use the molar mass of NH4Cl to convert moles to grams to find the mass produced.

Answer 2

The balanced chemical equation would be as follows:

NH3+HCL->NH4CL

For this, we assume these gases are ideal gases so we can use the equation PV=nRT. We first calculate the number of moles of each reactants. We do as follows:

PV=nRT
1.02 (4.21) = n (0.08206)(27+273.15)
n = 0.17 mol NH3 ------->consumed completely and therefore the limiting reactant

PV=nRT
0.998 (5.35 L) = n (0.08206)(26+273.15)
n = 0.22 mol HCl

what mass of NH4Cl(s) will be produced?
0.17 mol NH3 (1 mol NH4Cl / 1 mol NH3 ) = 0.17 mol NH3

which gas is the limiting reactant?
NH3 gas

which gas is present in excess?
HCl gas