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Ammonia reacts with copper(II) oxide. (i) Identify gas H. [1] (ii) Write a balanced chemical equation for the reaction in the combustion tube. [1] (iii) State two observations you would make in the co

Chemistry

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Answer

v

0.54 g\text{0.54 g}

Step-by-step solution

2 steps

Part i

Gas H is *Nitrogen

One use of nitrogen is to provide an inert atmosphere for packaging food or in light bulbs.

Part ii

The balanced chemical equation for the reaction in the combustion tube is

2NH3(g)+3CuO(s)→heat3Cu(s)+3H2O(l)+N2(g)\text{2NH}_3(g) + \text{3CuO}(s) \xrightarrow{\text{heat}} \text{3Cu}(s) + \text{3H}_2\text{O}(l) + \text{N}_2(g)

Part iii

Two observations in the combustion tube

  • The black solid (Copper(II) oxide) turns into a reddish-brown solid (Copper).
  • Colourless liquid droplets (water) are formed on the cooler parts of the tube.

Part iv

The colour of the contents in the water trough would change from *green to blue/purple

This is because any unreacted ammonia gas passing through the system would dissolve in the water, forming an alkaline solution. Universal indicator turns blue/purple in alkaline conditions.

Part v

0.54 g\text{0.54 g}

Step 1: Calculate the moles of ammonia.

Assuming molar volume of gas at room temperature and pressure (RTP) is 24000 cm3/mol24000 \text{ cm}^3/\text{mol}.

Moles of NH3=Volume of NH3Molar volume at RTP=480 cm324000 cm3/mol=0.02 mol\text{Moles of NH}_3 = \frac{\text{Volume of NH}_3}{\text{Molar volume at RTP}} = \frac{480 \text{ cm}^3}{24000 \text{ cm}^3/\text{mol}} = 0.02 \text{ mol}

  1. Use the mole ratio from the balanced equation to find moles of water (liquid Z).

    From the equation 2NH3(g)+3CuO(s)→3Cu(s)+3H2O(l)+N2(g)\text{2NH}_3(g) + \text{3CuO}(s) \rightarrow \text{3Cu}(s) + \text{3H}_2\text{O}(l) + \text{N}_2(g), the mole ratio of NH3:H2O\text{NH}_3 : \text{H}_2\text{O} is 2:32:3.

    Moles of H2O=Moles of NH3×32=0.02 mol×32=0.03 mol\text{Moles of H}_2\text{O} = \text{Moles of NH}_3 \times \frac{3}{2} = 0.02 \text{ mol} \times \frac{3}{2} = 0.03 \text{ mol}

  2. Calculate the mass of water (liquid Z).

    Molar mass of H2O=(2×1.0)+16.0=18.0 g/mol\text{H}_2\text{O} = (2 \times 1.0) + 16.0 = 18.0 \text{ g/mol}.

    Mass of H2O=Moles of H2O×Molar mass of H2O=0.03 mol×18.0 g/mol=0.54 g\text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times \text{Molar mass of H}_2\text{O} = 0.03 \text{ mol} \times 18.0 \text{ g/mol} = 0.54 \text{ g}

    The mass of liquid Z formed is 0.54 g\boxed{\text{0.54 g}}.

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