A chemist makes 86.94 grams of a black powder by reacting 54.94 grams of manganese with 32.00 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 find out how much manganese and oxygen can be obtained from 144.9 grams of the black powder, we first need to determine the composition of the black powder based on the original reaction.

1. Original Masses:

   • Manganese (Mn): 54.94 g
   • Oxygen (O): 32.00 g
   • Total mass of black powder produced: 86.94 g

2. Calculate the mass percentages of manganese and oxygen in the black powder: \[ \text{Mass percentage of Mn} = \left( \frac{54.94 \text{ g}}{86.94 \text{ g}} \right) \times 100 \approx 63.3% \] \[ \text{Mass percentage of O} = \left( \frac{32.00 \text{ g}}{86.94 \text{ g}} \right) \times 100 \approx 36.7% \]

3. Calculate the mass of manganese and oxygen in 144.9 grams of the black powder:

   • Mass of manganese in 144.9 g: \[ \text{Mass of Mn} = 0.633 \times 144.9 \text{ g} \approx 20.9 \text{ g} \]
   • Mass of oxygen in 144.9 g: \[ \text{Mass of O} = 0.367 \times 144.9 \text{ g} \approx 24.9 \text{ g} \]

So, if a student breaks down 144.9 grams of the black powder, they will obtain approximately:

• Mass of manganese: 20.9 grams
• Mass of oxygen: 24.9 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 as the black powder?

Answer 3

To determine whether the powder produced in this experiment is the same as the original black powder created from the reaction of 54.94 grams of manganese with 32.00 grams of oxygen, we need to calculate the composition of the new powder and compare it with the original powder.

Step 1: Determine the Composition of the New Powder

1. Original masses used in the experiment:

   • Manganese (Mn): 7.85 g
   • Oxygen (O): 2.29 g
   • Total mass of the powder produced: 10.14 g

2. Calculate the mass percentages of manganese and oxygen in the new powder:

   • Total mass of reactants = 7.85 g + 2.29 g = 10.14 g

   The reaction goes to completion since the total mass of the reactants equals the mass of the powder formed.

   • Mass percentage of Mn in the new powder: \[ \text{Mass percentage of Mn} = \left( \frac{7.85 \text{ g}}{10.14 \text{ g}} \right) \times 100 \approx 77.4% \]

   • Mass percentage of O in the new powder: \[ \text{Mass percentage of O} = \left( \frac{2.29 \text{ g}}{10.14 \text{ g}} \right) \times 100 \approx 22.6% \]

Step 2: Compare with the Original Black Powder Composition

From the previous calculation, the original black powder composition was approximately:

• Mass percentage of Mn: 63.3%
• Mass percentage of O: 36.7%

Conclusion:

The percentages of manganese and oxygen in the new powder (approximately 77.4% Mn and 22.6% O) differ significantly from those in the original black powder (approximately 63.3% Mn and 36.7% O).

Therefore, the powder produced in the new experiment is not the same as the original black powder. It has a different composition, suggesting it might be a different manganese oxide or compound formed due to the different amounts of reactants used.