Q1.
In an experiment, ammonia gas is made by heating a mixture of ammonium chloride and calcium
hydroxide.

2NH4Cl(s) + Ca(OH)2(s) =
CaCl2(s) + 2NH3(g) + 2H2O(l)

10.0 g of ammonium chloride is added to an excess of calcium hydroxide.

Calculate the maximum volume of ammonia gas that could be formed.

(relative atomic mass H = 1.00, N = 14.0, 0 = 16.0 and Ca = 40.0; one mole of any gas occupies 24 dm
at room temperature and pressure)
121

4.49dm3

Explanation:

2NH4Cl + Ca(OH)2 —> CaCl2 + 2NH3 + 2H2O

First, we need to convert 10g of ammonium chloride to mole. This is illustrated below:

Molar Mass of NH4Cl = 14 + (4x1) + 35.5 = 53.5g/mol

Mass of NH4Cl = 10g

Number of mole = Mass /Molar Mass

Number of mole of NH4Cl = 10/53.5 = 0.187mol

From the equation,

2moles of NH4Cl produced 2 moles of NH3.

Therefore, 0.187mol of NH4Cl will also produce 0.187mol of NH3

Now we can obtain the volume of NH3 produced by doing the following:

1mole of any gas occupy 24dm3

Therefore, 0.187mol of NH3 will occupy = 0.187 x 24 = 4.49dm3

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g