A chemist makes 86.94 grams of a black powder by reacting 54.94 grams of manganese with 32 grams of oxygen. If a student breaks down 144.9 grams of that powder, what mass of manganese and what mass of oxygen will be made?

Answer 1

To determine the mass of manganese and oxygen that can be produced from breaking down 144.9 grams of the black powder, we first need to identify the reaction that produced the black powder, as well as the mass ratios of the reactants involved.

From the information given:

1. Mass of manganese (Mn): 54.94 grams
2. Mass of oxygen (O): 32 grams
3. Total mass of black powder produced: 86.94 grams

This suggests that the black powder is likely manganese oxide (MnO or MnO₂, but we have to assume a specific oxide from the reaction). For this calculation, we'll assume the formation of manganese dioxide (MnO₂), which is produced from the reaction of manganese with oxygen:

\[ \text{Mn} + \text{O}_2 \rightarrow \text{MnO}_2 \]

Now we calculate the total mass:

1. The total mass of the reactants: \[ 54.94 , \text{g (Mn)} + 32 , \text{g (O)} = 86.94 , \text{g} \]

This confirms that the reaction yielded the 86.94 grams of black powder, which we assume to be MnO₂.

Now, we will calculate the mass proportions of manganese and oxygen in the black powder formed (MnO₂):

1. Molar mass of MnO₂:

   • Molar mass of Mn = 54.94 g/mol
   • Molar mass of O = 16.00 g/mol → For O₂, it's 32.00 g/mol; for O₂ it's 16.00 g/mol \[ \text{Molar mass of MnO}_2 = 54.94 + 2(16.00) = 54.94 + 32.00 = 86.94 , \text{g/mol} \]

2. Mass fractions:

   • Mass fraction of Mn in MnO₂ = \( \frac{54.94}{86.94} \)
   • Mass fraction of O in MnO₂ = \( \frac{32.00}{86.94} \)

Now we apply these mass fractions to the total mass of black powder (144.9 g) to find the masses of manganese and oxygen produced:

\[ \text{mass of Mn in 144.9 g of MnO}_2 = 144.9 , \text{g} \times \frac{54.94}{86.94} \approx 94.47 , \text{g} \]

\[ \text{mass of O in 144.9 g of MnO}_2 = 144.9 , \text{g} \times \frac{32.00}{86.94} \approx 50.43 , \text{g} \]

Thus, when a student breaks down 144.9 grams of the black powder, approximately:

• Mass of manganese: 94.47 grams
• Mass of oxygen: 50.43 grams

Answer 2

In an experiment, a chemist reacts 7.85 grams of manganese with 2.29 grams of oxygen to make 10.14 grams of a powder. Is it the same powder in the last problem?

Answer 3

To determine if the powder produced in this new experiment is the same as the manganese oxide powder discussed in the previous problem, we need to analyze the mass ratios of the reactants and the product.

Step 1: Calculate the moles of each reactant

1. Manganese (Mn):

   • Molar mass of manganese (Mn) = \( 54.94 , \text{g/mol} \) \[ \text{Moles of Mn} = \frac{7.85 , \text{g}}{54.94 , \text{g/mol}} \approx 0.143, \text{mol} \]

2. Oxygen (O):

   • Molar mass of oxygen (O) = \( 16.00 , \text{g/mol} \) \[ \text{Moles of O} = \frac{2.29 , \text{g}}{32.00 , \text{g/mol}} \approx 0.0716, \text{mol} \]

Step 2: Identify the stoichiometry of the reaction

The reaction for forming manganese dioxide can be represented as: \[ \text{2 Mn} + \text{O}_2 \rightarrow \text{2 MnO}_2 \]

From the reaction, we see that:

• 1 mole of O₂ reacts with 2 moles of Mn.

Step 3: Determine the mole ratio of Mn to O₂ in the current experiment

• The amount of manganese required for the 0.0716 moles of oxygen according to the stoichiometry: \[ \text{Moles of Mn needed} = 2 \times 0.0716 \approx 0.143, \text{mol} \]

The chemist has \( 0.143 , \text{mol} \) of manganese, precisely matching the requirement for the given amount of oxygen.

Step 4: Calculate the theoretical yield of MnO₂

Theoretical mass of the product (MnO₂):

1. Each mole of MnO₂ consists of 1 mole of Mn and 2 moles of O.

2. Therefore, the mass of MnO₂ produced can be calculated by the moles of Mn available (since Mn is the limiting reactant here, based on the reaction stoichiometry):

   • Total moles of MnO₂ produced: \[ \text{Moles of MnO}_2 = \text{Moles of Mn} \approx 0.143 , \text{mol} \]
   • Molar mass of MnO₂ = \( 54.94 + (2 \times 16.00) = 54.94 + 32.00 = 86.94 , \text{g/mol} \)

   Hence, the total mass of MnO₂ produced is: \[ \text{Mass of MnO}_2 = 0.143 , \text{mol} \times 86.94 , \text{g/mol} \approx 12.44 , \text{g} \]

Comparison with the experimental result

The actual yield in this case is 10.14 grams of powder. This mass is less than the theoretical mass (12.44 grams), suggesting that the reaction may not have gone to completion or there were side products.

Conclusion

Since the powder produced from this reaction contains manganese and oxygen in a consistent 1:2 ratio similar to manganese dioxide (MnO₂) and given the reaction conditions and stoichiometry are the same, it can be concluded that:

Yes, the powder produced in this experiment is likely manganese dioxide (MnO₂), the same powder as in the previous problem. The discrepancy in mass can be attributed to incomplete reaction or loss during the process.