Question

When ammonia is mixed with hydrogen chloride (HCl), the white solid ammonium chloride (NH4Cl) is produced. Suppose 10.0 g ammonia is mixed with the same mass of hydrogen chloride. What substances will be present after the reaction has gone to completion, and what will their masses be?

Answer 1

There will remain 5.33 grams of NH3 and there will be produced 14.66 grams NH4Cl

Step-by-step explanation:

Step 1: Data given

Mass of ammonia = 10.0 grams

Mass of hydrogen chloride = 10.0 grams

Molar mass of ammonia = 17.03 g/mol

Molar mass of hydrogen chloride = 36.46 g/mol

Step 2: the balanced equation

NH3 + HCl → NH4Cl

Step 3: Calculate moles NH3

Moles NH3 = mass NH3 / molar mass NH3

Moles NH3 = 10.0 grams / 17.03 g/mol

Moles NH3 = 0.587 moles

Step 4: Calculate moles HCl

Moles HCl = 10.0 grams / 36.46 g/mol

Moles HCl = 0.274 moles

Step 5: Calculate limiting reactant

For 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4Cl

HCl is the limiting reactant. It will completely be consumed. (0.274 moles)

NH3 is in excess. There will remain 0.587 - 0.274 = 0.313 moles

This is: 0.313 moles * 17.03 = 5.33 grams

Step 6: Calculate moles NH4Cl

For 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4Cl

For 0.274 moles HCl we need 0.274 moles NH4Cl

Step 7: Calculate mass of NH4Cl

Mass NH4Cl = moles NH4Cl * molar mass NH4Cl

Mass NH4Cl = 0.274 moles * 53.49 g/mol

Mass NH4Cl = 14.66 grams

There will remain 5.33 grams of NH3 and there will be produced 14.66 grams NH4Cl