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An indigestion tablet contains a mass of 0.30 g of magnesium hydroxide [Mg(OH)2] as its only basic ingredient - Leaving Cert Chemistry - Question a - 2005

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An indigestion tablet contains a mass of 0.30 g of magnesium hydroxide [Mg(OH)2] as its only basic ingredient. The balanced chemical equation for the reaction betwee... show full transcript

Worked Solution & Example Answer:An indigestion tablet contains a mass of 0.30 g of magnesium hydroxide [Mg(OH)2] as its only basic ingredient - Leaving Cert Chemistry - Question a - 2005

Step 1

Calculate the volume of 1.0 M HCl neutralised by two of these indigestion tablets.

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Answer

To find the volume of HCl neutralised:

  1. Calculate moles of Mg(OH)2 in two tablets:

    • For one tablet: 0.30extg×1extmol58.3extg=0.00513 mol≈0.0052extmol0.30 ext{ g} \times \frac{1 ext{ mol}}{58.3 ext{ g}} = 0.00513 \text{ mol} \approx 0.0052 ext{ mol}
    • For two tablets: 0.0052 mol×2=0.0104 mol0.0052 \text{ mol} \times 2 = 0.0104 \text{ mol}

  2. Using the reaction stoichiometry, moles of HCl required:

    • From the reaction: 1 mol Mg(OH)2 reacts with 2 mol HCl
    • Therefore: 0.0104 mol Mg(OH)2×2 mol HCl=0.0208 mol HCl0.0104 \text{ mol Mg(OH)}_2 \times 2 \text{ mol HCl} = 0.0208 \text{ mol HCl}

  3. Calculate volume of 1.0 M HCl required: extVolume=0.0208 mol1.0 M=0.0208 L=20.8 cm3≈21 cm3 ext{Volume} = \frac{0.0208 \text{ mol}}{1.0 \text{ M}} = 0.0208 \text{ L} = 20.8 \text{ cm}^3 \approx 21 \text{ cm}^3

Step 2

What mass of salt is formed in this neutralisation?

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Answer

To calculate the mass of salt formed:

  1. Determine moles of MgCl2 formed:

    • From the stoichiometry: 1 mol of Mg(OH)2 produces 1 mol of MgCl2
    • Therefore, moles of MgCl2: 0.0104 mol0.0104 \text{ mol}

  2. Calculate mass of MgCl2:

    • Molar mass of MgCl2: 24.3extg/mol(Mg)+2×35.5extg/mol(Cl)=95.3extg/mol24.3 ext{ g/mol (Mg)} + 2 \times 35.5 ext{ g/mol (Cl)} = 95.3 ext{ g/mol}
    • Thus, mass: 0.0104 mol×95.3extg/mol≈0.99extg0.0104 \text{ mol} \times 95.3 ext{ g/mol} \approx 0.99 ext{ g}

Step 3

How many magnesium ions are present in this amount of the salt?

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Answer

To find the number of magnesium ions in MgCl2:

  1. Moles of Mg ions in MgCl2:

    • For every 1 mol of MgCl2, there is 1 mol of Mg ions.
    • Therefore: 0.0104 mol of MgCl2=0.0104 mol of Mg2+0.0104 \text{ mol of MgCl}_2 = 0.0104 \text{ mol of Mg}^{2+}

  2. Calculate number of Mg ions:

    • Using Avogadro's number (approximately 6.02×10236.02 \times 10^{23}): 0.0104 mol×6.02×1023 ions/mol≈6.26×1021 Mg2+ ions0.0104 \text{ mol} \times 6.02 \times 10^{23} \text{ ions/mol} \approx 6.26 \times 10^{21} \text{ Mg}^{2+} \text{ ions}

Step 4

What volume of this second indigestion tablet remedy would have the same neutralising effect?

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Answer

To find the volume of the second remedy:

  1. Start with the amount of Mg(OH)2 that neutralises equivalent stomach acid:

    • As previously calculated, 0.30 g of Mg(OH)2 allows a neutralisation of 0.0104 mol.

  2. For a 6% (w/v) solution:

    • This means 6 g of Mg(OH)2 is present in 100 cm³ of solution.

  3. Find how much solution is required for 0.30 g:

    • The mass of Mg(OH)2 in 10 cm³: 6extg100extcm3=0.06extg/cm3\frac{6 ext{ g}}{100 ext{ cm}^3} = 0.06 ext{ g/cm}^3
    • Therefore, for 0.30 g: Volume=0.30extg0.06extg/cm3=5extcm3\text{Volume} = \frac{0.30 ext{ g}}{0.06 ext{ g/cm}^3} = 5 ext{ cm}^3
    • Hence, the required volume is 10 cm³.

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