Question

chemistry
unit 6
stoichiometry
homework / notes packet
limiting reactants warm up
tyler casing p.1
1.) magnesium and oxygen react to form magnesium oxide and magnesium peroxide
2.) if you have 12.0 grams of magnesium and 16.0 grams of oxygen:

1. what is the limiting reactant?
2. how many grams of magnesium peroxide can you make?

Explanation:

First, we define the balanced chemical equation for the formation of magnesium peroxide ($\text{MgO}_2$):

$$2\text{Mg} + 2\text{O}_2 ightarrow 2\text{MgO}_2$$

Simplified to:

$$\text{Mg} + \text{O}_2 ightarrow \text{MgO}_2$$

Step1: Calculate moles of Mg

Molar mass of $\text{Mg} = 24.305\ \text{g/mol}$
$$n_{\text{Mg}} = \frac{12.0\ \text{g}}{24.305\ \text{g/mol}} \approx 0.494\ \text{mol}$$

Step2: Calculate moles of $\text{O}_2$

Molar mass of $\text{O}_2 = 32.00\ \text{g/mol}$
$$n_{\text{O}_2} = \frac{16.0\ \text{g}}{32.00\ \text{g/mol}} = 0.500\ \text{mol}$$

Step3: Identify limiting reactant

From the balanced equation, 1 mol $\text{Mg}$ reacts with 1 mol $\text{O}_2$. Moles of $\text{Mg}$ ($0.494\ \text{mol}$) < moles of $\text{O}_2$ ($0.500\ \text{mol}$), so $\text{Mg}$ is limiting.

Step4: Calculate moles of $\text{MgO}_2$

Moles of $\text{MgO}_2$ = moles of limiting $\text{Mg} = 0.494\ \text{mol}$

Step5: Calculate mass of $\text{MgO}_2$

Molar mass of $\text{MgO}_2 = 24.305 + 2\times16.00 = 56.305\ \text{g/mol}$
$$m_{\text{MgO}_2} = 0.494\ \text{mol} \times 56.305\ \text{g/mol} \approx 27.8\ \text{g}$$

Answer:

1. Limiting reactant: Magnesium ($\text{Mg}$)
2. Mass of magnesium peroxide: $\approx 27.8$ grams