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Problem: Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid ammonium chloride, NH4Cl(s):NH3(g) + HCl(g) → NH4Cl(s)Two 2.50 L flasks at 30.0 oC are connected by a stopcock, as shown in the drawingOne flask contains 5.60g NH3(g), and the other contains 4.60 g  HCl(g). When the stopcock is opened, the gases react until one is completely consumed.What mass of ammonium chloride will be formed?

FREE Expert Solution

Step 1

5.60 g NH3 ×1 mol NH317.031 g NH3×1 mol NH4Cl1 mol NH3 = 0.3288 mol NH4Cl

4.60 g HCl ×1 mol HCl36.46 g HCl×1 mol NH4Cl1 mol HCl = 0.1262 mol NH4Cl


HCl → limiting reactant


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Problem Details

Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid ammonium chloride, NH4Cl(s):
NH3(g) + HCl(g) → NH4Cl(s)
Two 2.50 L flasks at 30.0 oC are connected by a stopcock, as shown in the drawing

Two gas flasks are connected by a tube with a stopcock. NH3 is in the flask on the left and HCl is in the flask on the right.

One flask contains 5.60g NH3(g), and the other contains 4.60 g  HCl(g). When the stopcock is opened, the gases react until one is completely consumed.

What mass of ammonium chloride will be formed?

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Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.