Question

The set up was used by a student to pass dry ammonia gas over heated vopper (ii) oxide(a)Give two observations made in the combustion tube (b)Name a suitable drying agent (c)On the experiment the student passed dry ammonia over 565g of copper (ii)oxide until the reaction was complete (i)Write an equation for the reaction that took place (ii)What property of ammonia is shown by the experiment? (iii)Calculate the mass of the copper produced.(Cu=63.5,O=16) (iv)Calculate the volume of the gas produced at s.t.p. (Molar gas volume at s.t.p 22.4dm^3)

Answer 1

A suitable drying agent for ammonia gas is magnesium perchlorate, and the reaction between ammonia and copper (II) oxide demonstrates ammonia's reducing property. The equations and calculations provided show the mass of copper and volume of nitrogen gas produced in this reaction.

Step-by-step explanation:

When dry ammonia gas is passed over heated copper (II) oxide, two observations in the combustion tube can be seen:

A suitable drying agent for ammonia gas is magnesium perchlorate (Mg(CIO4)2).

The equation for the reaction that takes place is:

3CuO + 2NH3 -> 3Cu + N2 + 3H2O

This experiment showcases ammonia's reducing property. To calculate the mass of copper produced:

The molar mass of CuO = 63.5 (Cu) + 16 (O) = 79.5 g/mol

The number of moles of CuO = 565 g / 79.5 g/mol = 7.11 mole

Since the reaction has a 3:3 mole ratio between CuO and Cu:

The mass of copper produced = 7.11 moles * 63.5 g/mol (Cu) = 451.485 g

To calculate the volume of nitrogen gas produced:

The mole ratio between CuO and N2 is 3:1, so moles of N2 = 7.11 moles / 3 = 2.37 moles

The volume of nitrogen gas at s.t.p. = 2.37 moles * 22.4 dm^3/mol = 53.088 dm^3