If 150g of impure ribbon reacted with excess hydrochloric acid liberating 8.96dm3 of hydrogen gas at s.t.p

Question

Calculate the percentage purity of the magnesium ribbon
Number of chlorine ions produced

Answer 1

To determine the percentage purity of the magnesium ribbon and the number of chlorine ions produced, we can use the information provided about the reaction of magnesium with hydrochloric acid. The relevant chemical reaction is:

$$\text{Mg (s)}+2\text{HCl (aq)}\to {\text{MgCl}}_2\text{(aq)}+{\text{H}}_2\text{(g)}$$

1. Calculate the number of moles of hydrogen gas produced.

At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 dm³. We are given that 8.96 dm³ of hydrogen gas was produced:

$$\text{Number of moles of }{H}_2=\frac{\text{Volume of }{H}_2}{22.4{\text{ dm}}^3/\text{mol}}=\frac{8.96{\text{ dm}}^3}{22.4{\text{ dm}}^3/\text{mol}}\approx 0.400\text{ moles }{H}_2$$

2. Calculate the number of moles of magnesium that reacted.

From the balanced chemical equation, 1 mole of magnesium reacts with 1 mole of hydrogen gas. Therefore, the number of moles of magnesium that reacted is also:

$$\text{Number of moles of Mg}=0.400\text{ moles}$$

3. Calculate the mass of magnesium that reacted.

The molar mass of magnesium (Mg) is approximately 24.31 g/mol. Hence, the mass of magnesium that reacted can be calculated as follows:

$$\text{Mass of }Mg=\text{Number of moles of }Mg\times \text{Molar mass of }Mg=0.400\text{ moles}\times 24.31\text{ g/mol}\approx 9.724\text{ g}$$

4. Calculate the percentage purity of the magnesium ribbon.

Given that the total mass of the impure magnesium ribbon is 150 g, the percentage purity can be calculated as follows:

$$\text{Percentage purity}=\left(\frac{\text{Mass of pure }Mg}{\text{Total mass of impure ribbon}}\right)\times 100$$

$$\text{Percentage purity}=\left(\frac{9.724\text{ g}}{150\text{ g}}\right)\times 100\approx 6.82$$

5. Calculate the number of chlorine ions produced.

From the balanced equation, 1 mole of Mg produces 1 mole of MgCl₂, which contains 2 moles of Cl⁻ ions. Therefore, the number of moles of Cl⁻ ions produced is:

$$\text{Number of moles of }C{l}^{-}=2\times \text{Number of moles of }Mg=2\times 0.400\approx 0.800{\text{ moles of Cl}}^{-}$$

Since 1 mole of Cl⁻ ions contains Avogadro's number of entities (approximately $6.022\times {10}^{23}$), the total number of chlorine ions produced can be calculated as:

$${\text{Number of Cl}}^{-}\text{ ions}=0.800\text{ moles}\times 6.022\times {10}^{23}\text{ ions/mole}\approx 4.816\times {10}^{23}\text{ ions}$$

Summary of Results:

• Percentage purity of magnesium ribbon: 6.82%
• Number of chlorine ions produced: approximately $4.82\times {10}^{23}$ ions