Question

As part of a laboratory experiment, ammonia gas was generated according to the equation shown below. 21.4 g of ammonium chloride and 22.2 g of calcium hydroxide were used in the reaction, use this information to calculate the mass of ammonia gas produced in grams.

-Give your answer to one decimal place

-You do not need to quote the units in your answer

2 NH4Cl + Ca(OH)2 → CaCl2 + 2 NH3 + 2 H2O

Answer 1

2NH₄ Cl + Ca(OH)₂ -> CacL₂ + 2NH₃ + 2H₂O

Mass of ammonium chloride = 53.5g

Moles = mass/ molar mass = 21.4/ 53.5 = 0.4

Mass of Calcium hydroxide = 22.2 g

Molar mass of Calcium hydroxide = 74 g

Moles of calcium hydroxide = 22.2/74 = 0.3

Limiting reagent has minimum moles/coefficient ratio

For NH₄Cl moles/coeff = 0.4/2 = 0.2

For C_a(OH)₂ moles/coeff = 0.3/1 = 0.3

So NH_⁴Cl is limiting reagent

Z moles NH⁴Cl produce Z moles NH³

0.4 moles NH⁴Cl produce 0.4 moles NH³

mass of ammonia produce 0.4 x 17 g [molar mass NH³ = 17

= 6.8g