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