A sample of calcium carbide of mass 2.5g reacts with water to produce acetylene gas and calcium hydroxide. i) State the formula of acetylene. ii) Write the balanced chemical equation for the reaction.

Here are the solutions to your homework questions:

5.

a)

• (i) The gas obtained is acetylene, and its formula is $\boxed{\text{C}_2\text{H}_2}$.

• (ii) The balanced chemical equation is:

  $$\boxed{\text{CaC}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{C}_2\text{H}_2(g)}$$

• (iii) Calculate the mass of slaked lime obtained from 2.5g of calcium carbide.

  Step 1: Calculate the molar mass of $\text{CaC}_2$ and $\text{Ca(OH)}_2$.
  Molar mass of $\text{CaC}_2 = 40 + (2 \times 12) = 64 \text{ g/mol}$.
  Molar mass of $\text{Ca(OH)}_2 = 40 + (2 \times 16) + (2 \times 1) = 74 \text{ g/mol}$.

  Step 2: Calculate the moles of $\text{CaC}_2$ used.

  $$\text{Moles of CaC}_2 = \frac{2.5 \text{ g}}{64 \text{ g/mol}} = 0.0390625 \text{ mol}$$

  Step 3: From the balanced equation, 1 mole of $\text{CaC}_2$ produces 1 mole of $\text{Ca(OH)}_2$. So, moles of $\text{Ca(OH)}_2$ produced = $0.0390625 \text{ mol}$.

  Step 4: Calculate the mass of $\text{Ca(OH)}_2$.

  $$\text{Mass of Ca(OH)}_2 = 0.0390625 \text{ mol} \times 74 \text{ g/mol} = 2.890625 \text{ g}$$

  The mass of slaked lime obtained is $\boxed{\text{2.89 g}}$.

b)

• (i) The number of Faradays used in this experiment.

  Step 1: Calculate the total charge passed.

  $$Q = I \times t = 2 \text{ A} \times (15 \times 60) \text{ s} = 2 \text{ A} \times 900 \text{ s} = 1800 \text{ C}$$

  Step 2: Calculate the number of Faradays. (Using Faraday's constant $F = 96485 \text{ C/mol}$)

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