Question

Consider the following reaction and based on the reaction answer the questions that follow:

Calculate:

1) the quantity in moles of (NH₄)₂Cr₂O₇ if 63gm of(NH₄)₂Cr₂O₇ is heated.

2) the quantity in moles of nitrogen formed.

3) the volume in liters or dm3 of N2 evolved at S.T.P.

4) the mass in grams of Cr2O3 formed at the same time

(Atomic masses: H=1, Cr= 52, N=14]

Answer 1

The given reaction is as follows:

1) Given:

Weight of (NH₄)₂Cr₂O₇ = 63 gm

Molar mass of (NH₄)₂Cr₂O₇

= (2 x 14) + (8 x 1) + (2 x 52) + (7 x 16)

= 28 + 8 + 104 + 112

= 252 gm

1 mole (NH₄)₂Cr₂O₇ = 252 gm

Hence, 63 gm of (NH₄)₂Cr₂O₇ = 0.25 moles

The quantity of moles of (NH₄)₂Cr₂O₇ if 63 gm of (NH₄)₂Cr₂O₇ is heated is 0.25 moles.

2) From the given chemical equation, 1 mole of (NH₄)₂Cr₂O₇ produces 1 mole of nitrogen gas.

Hence, 0.25 moles of (NH₄)₂Cr₂O₇ can produce 0.25 moles of nitrogen gas.

The quantity in moles of nitrogen formed is 0.25 moles

3) One mole of an ideal gas at S.T.P. occupies 22.4 liters or dm³.

Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will occupy 0.25  22.4 = 5.6 litres or dm³

The volume in liters or dm3 of N2 evolved at S.T.P. is 5.6 liters or dm³.

4)From the given chemical equation, 1 mole of (NH₄)₂Cr₂O₇ produces 1 mole of Cr₂O₃.

Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will produce 0.25 moles of Cr₂O₃

Molar mass of Cr₂O₃

= (2 x 52) + (3 x 16)

= 104 + 48

= 152 gm

1 mole Cr₂O₃ = 152 gm

Hence, 0.25 moles of Cr₂O₃ = 0.25  152 = 38 gm

The mass in grams of Cr₂O₃ formed at the same time is 38 gm.