Questions: Balance the chemical equation given below, and determine the number of grams of MgO needed to produce 15.0 g of Fe2 O3. MgO(s)+ Fe(s) -> Fe2 O3(s)+ Mg(s)

Solution Steps

To balance the chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The unbalanced equation is:

\[ \mathrm{MgO}(s) + \mathrm{Fe}(s) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s) + \mathrm{Mg}(s) \]

Balancing the equation involves the following steps:

1. Balance the iron (Fe) atoms: We have 2 Fe atoms in \(\mathrm{Fe}_{2}\mathrm{O}_{3}\), so we need 2 Fe atoms on the reactant side.
2. Balance the magnesium (Mg) atoms: We have 1 Mg atom in \(\mathrm{MgO}\) and 1 Mg atom in \(\mathrm{Mg}\), so they are already balanced.
3. Balance the oxygen (O) atoms: We have 3 O atoms in \(\mathrm{Fe}_{2}\mathrm{O}_{3}\), so we need 3 \(\mathrm{MgO}\) molecules to provide 3 O atoms.

The balanced equation is:

\[ 3\mathrm{MgO}(s) + 2\mathrm{Fe}(s) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s) + 3\mathrm{Mg}(s) \]

Next, we calculate the molar masses of the compounds involved:

• Molar mass of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\): \[ 2 \times 55.845 + 3 \times 15.999 = 159.687 \, \text{g/mol} \]

• Molar mass of \(\mathrm{MgO}\): \[ 24.305 + 15.999 = 40.304 \, \text{g/mol} \]

Calculate the moles of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) produced from 15.0 g:

\[ \text{Moles of } \mathrm{Fe}_{2}\mathrm{O}_{3} = \frac{15.0 \, \text{g}}{159.687 \, \text{g/mol}} = 0.0939 \, \text{mol} \]

From the balanced equation, 1 mole of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) is produced from 3 moles of \(\mathrm{MgO}\). Therefore, the moles of \(\mathrm{MgO}\) needed are:

\[ \text{Moles of } \mathrm{MgO} = 0.0939 \, \text{mol} \times 3 = 0.2817 \, \text{mol} \]

Finally, calculate the mass of \(\mathrm{MgO}\) required:

\[ \text{Mass of } \mathrm{MgO} = 0.2817 \, \text{mol} \times 40.304 \, \text{g/mol} = 11.35 \, \text{g} \]

Final Answer