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Identify gas H and explain its use in food packaging. Write the balanced chemical equation for the reaction. State two observations and explain the colour change in the water trough. Calculate the mas

Chemistry

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Answer

a(v)

0.54 g\text{0.54 g}

Step-by-step solution

3 steps

Part a(i)

Gas H is nitrogen (N2\text{N}_2).
One use of nitrogen is to provide an inert atmosphere for packaging food.

Part a(ii)

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

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

Part a(iii)

Two observations in the combustion tube as the experiment progresses are:

  • The black solid (copper(II) oxide) turns into a brown/pinkish-brown solid (copper metal).
  • Colourless liquid droplets (water) are observed in the cooler parts of the combustion tube or the connecting tube.

Part a(iv)

The colour of the contents in the water trough changed from green to blue/purple.
This is because unreacted ammonia gas, which is alkaline, dissolves in the water to form ammonium hydroxide. This alkaline solution turns the universal indicator blue or purple.

Part a(v)

0.54 g\text{0.54 g}

To calculate the mass of liquid Z (water) formed:

  1. Calculate the moles of ammonia (NH3\text{NH}_3) that reacted.

    We assume the reaction occurs at Room Temperature and Pressure (RTP), where the molar volume of a gas is 24000 cm3/mol24000 \text{ cm}^3/\text{mol}.

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

  2. Use the stoichiometry from the balanced equation to find the moles of water (H2O\text{H}_2\text{O}) produced.

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

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

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

    The molar mass of H2O\text{H}_2\text{O} is 2(1.0)+16.0=18.0 g/mol2(1.0) + 16.0 = 18.0 \text{ g/mol}.

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