Questions: A typical bottle of "Milk of Magnesia" contains 75 grams of MAGNESIUM HYDROXIDE. Show how to calculate the mass of MAGNESIUM SULFATE and the mass of SODIUM HYDROXIDE needed to make 75 grams of Mg(OH)2 ? Hint: each answer requires a separate proportion!

A typical bottle of "Milk of Magnesia" contains 75 grams of MAGNESIUM HYDROXIDE. Show how to calculate the mass of MAGNESIUM SULFATE and the mass of SODIUM HYDROXIDE needed to make 75 grams of Mg(OH)2 ? Hint: each answer requires a separate proportion!
Transcript text: A typical bottle of "Milk of Magnesia" contains 75 grams of MAGNESIUM HYDROXIDE. Show how to calculate the mass of MAGNESIUM SULFATE and the mass of SODIUM HYDROXIDE needed to make 75 grams of $\mathrm{Mg}(\mathrm{OH})_{2}$ ? Hint: each answer requires a separate proportion!
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Solution

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Solution Steps

Step 1: Determine the Molar Mass of Magnesium Hydroxide

First, we need to calculate the molar mass of magnesium hydroxide, \(\mathrm{Mg(OH)_2}\).

  • Magnesium (Mg): \(24.305 \, \text{g/mol}\)
  • Oxygen (O): \(16.00 \, \text{g/mol}\)
  • Hydrogen (H): \(1.008 \, \text{g/mol}\)

The molar mass of \(\mathrm{Mg(OH)_2}\) is calculated as follows:

\[ \text{Molar mass of } \mathrm{Mg(OH)_2} = 24.305 + 2 \times (16.00 + 1.008) = 58.319 \, \text{g/mol} \]

Step 2: Calculate the Moles of Magnesium Hydroxide

Next, we calculate the number of moles of \(\mathrm{Mg(OH)_2}\) in 75 grams.

\[ \text{Moles of } \mathrm{Mg(OH)_2} = \frac{75 \, \text{g}}{58.319 \, \text{g/mol}} = 1.286 \, \text{mol} \]

Step 3: Calculate the Mass of Magnesium Sulfate Needed

Magnesium sulfate (\(\mathrm{MgSO_4}\)) is used to produce magnesium hydroxide. We need to find the mass of \(\mathrm{MgSO_4}\) required to produce 1.286 moles of \(\mathrm{Mg(OH)_2}\).

The molar mass of \(\mathrm{MgSO_4}\) is:

  • Magnesium (Mg): \(24.305 \, \text{g/mol}\)
  • Sulfur (S): \(32.07 \, \text{g/mol}\)
  • Oxygen (O): \(16.00 \, \text{g/mol} \times 4\)

\[ \text{Molar mass of } \mathrm{MgSO_4} = 24.305 + 32.07 + 4 \times 16.00 = 120.375 \, \text{g/mol} \]

The mass of \(\mathrm{MgSO_4}\) needed is:

\[ \text{Mass of } \mathrm{MgSO_4} = 1.286 \, \text{mol} \times 120.375 \, \text{g/mol} = 154.8 \, \text{g} \]

Step 4: Calculate the Mass of Sodium Hydroxide Needed

Sodium hydroxide (\(\mathrm{NaOH}\)) is also used in the reaction. We need to find the mass of \(\mathrm{NaOH}\) required to produce 1.286 moles of \(\mathrm{Mg(OH)_2}\).

The molar mass of \(\mathrm{NaOH}\) is:

  • Sodium (Na): \(22.99 \, \text{g/mol}\)
  • Oxygen (O): \(16.00 \, \text{g/mol}\)
  • Hydrogen (H): \(1.008 \, \text{g/mol}\)

\[ \text{Molar mass of } \mathrm{NaOH} = 22.99 + 16.00 + 1.008 = 39.998 \, \text{g/mol} \]

The mass of \(\mathrm{NaOH}\) needed is:

\[ \text{Mass of } \mathrm{NaOH} = 1.286 \, \text{mol} \times 39.998 \, \text{g/mol} = 51.44 \, \text{g} \]

Final Answer

\[ \boxed{\text{Mass of } \mathrm{MgSO_4} = 154.8 \, \text{g}} \] \[ \boxed{\text{Mass of } \mathrm{NaOH} = 51.44 \, \text{g}} \]

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