Question

Experiment 4:%0D%0A%0D%0A%0D%0A%0D%0A(1 point)%0D%0A7. Predict the mass of oxygen that will be left over after the reaction of 45.7 g of magnesium and 55.2 g of oxygen.%0D%0A%0D%0AMass of left over oxygen in the products:

Answer 1

the answer options are
21.4g
45.7g
55.2g
79.5g
100.9g

Answer 2

To predict the mass of oxygen left over after the reaction of magnesium and oxygen, we first need to look at the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O₂):

\[
2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO}
\]

### Step 1: Calculate moles of reactants

#### Moles of magnesium (Mg):
- Molar mass of Mg = 24.31 g/mol
- Moles of Mg = mass / molar mass = \( \frac{45.7 \text{ g}}{24.31 \text{ g/mol}} \approx 1.88 \text{ mol} \)

#### Moles of oxygen (O₂):
- Molar mass of O₂ = 32.00 g/mol
- Moles of O₂ = mass / molar mass = \( \frac{55.2 \text{ g}}{32.00 \text{ g/mol}} \approx 1.73 \text{ mol} \)

### Step 2: Determine the limiting reactant

From the balanced equation, we can see that:
- 2 moles of Mg reacts with 1 mole of O₂.
- This means that 1 mole of O₂ requires 2 moles of Mg.

#### Required moles of Mg for 1.73 moles of O₂:
\[
\text{Required moles of Mg} = 1.73 \text{ mol} \times 2 = 3.46 \text{ mol}
\]

Since we only have 1.88 moles of Mg available, magnesium is the limiting reactant.

### Step 3: Calculate the amount of O₂ reacted

From the stoichiometry of the reaction, 2 moles of Mg consume 1 mole of O₂.

For 1.88 moles of Mg:
\[
\text{Moles of O₂ reacted} = \frac{1.88 \text{ mol Mg}}{2} = 0.94 \text{ mol O₂}
\]

### Step 4: Calculate the remaining moles of O₂

Initially, we had 1.73 moles of O₂. After the reaction, the remaining moles of O₂ will be:
\[
\text{Remaining moles of O₂} = 1.73 \text{ mol} - 0.94 \text{ mol} = 0.79 \text{ mol}
\]

### Step 5: Calculate the mass of leftover O₂

Now, we need to convert the remaining moles of O₂ back to grams:
\[
\text{Mass of leftover O₂} = \text{moles} \times \text{molar mass} = 0.79 \text{ mol} \times 32.00 \text{ g/mol} = 25.28 \text{ g}
\]

### Final Answer

The mass of oxygen that will be left over after the reaction is **25.28 g**.

Answer 3

Let's recalculate the mass of leftover oxygen to ensure we identify the correct answer among your provided options.

**Balanced Chemical Equation:**
\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

### Step 1: Calculate moles of reactants

#### Moles of magnesium (Mg):
- Molar mass of Mg = 24.31 g/mol
- Moles of Mg = \( \frac{45.7 \text{ g}}{24.31 \text{ g/mol}} \approx 1.88 \, \text{mol} \)

#### Moles of oxygen (O₂):
- Molar mass of O₂ = 32.00 g/mol
- Moles of O₂ = \( \frac{55.2 \text{ g}}{32.00 \text{ g/mol}} \approx 1.73 \, \text{mol} \)

### Step 2: Determine the limiting reactant

According to the balanced equation:
- 2 moles of Mg react with 1 mole of O₂, meaning 1 mole of O₂ requires 2 moles of Mg.

#### Required moles of Mg for 1.73 moles of O₂:
\[
\text{Required moles of Mg} = 1.73 \, \text{mol} \times 2 = 3.46 \, \text{mol}
\]

Since we only have 1.88 moles of Mg, Mg is the limiting reactant.

### Step 3: Calculate the amount of O₂ that reacts

From the stoichiometry of the reaction:
- 2 moles of Mg react with 1 mole of O₂.

For 1.88 moles of Mg:
\[
\text{Moles of O₂ reacted} = \frac{1.88 \, \text{mol Mg}}{2} = 0.94 \, \text{mol O₂}
\]

### Step 4: Calculate the remaining moles of O₂

Initially, we have 1.73 moles of O₂. After the reaction, the remaining moles of O₂:
\[
\text{Remaining moles of O₂} = 1.73 \, \text{mol} - 0.94 \, \text{mol} = 0.79 \, \text{mol}
\]

### Step 5: Convert remaining moles of O₂ to grams

Using the molar mass of O₂ (32.00 g/mol):
\[
\text{Mass of leftover O₂} = 0.79 \, \text{mol} \times 32.00 \, \text{g/mol} = 25.28 \, \text{g}
\]

Since 25.28 g is not one of the provided options, let's check the calculations again considering the potential rounding issues or other errors.

### Re-evaluate Moles of O₂:

- If we initially have 1.73 moles of O₂ and reacted 0.94 moles:
- Remaining O₂ = 1.73 - 0.94 = 0.79 moles
- Mass remaining = 0.79 moles × 32.00 g/mol = 25.28 g (which aligns with our previous calculation).

However, given that 25.28 g is not a choice, let's consider if there's a mistake concerning the limiting reactants or their stoichiometry.

### Final Check:

Reassessing options:
Out of the options, **21.4g** is the closest option but does not precisely match our calculation. The calculations consistently lead to 25.28 g.

The discrepancy might be due to rounding, but since 25.28 g is not listed, there isn't a clear choice from the current options based on standard stoichiometry.

Therefore, if you have to choose based on what's closest, **21.4 g** seems the most reasonable, however, please double-check the inputs or constraints of the experiment to align results with expected outputs.