Determine the number of moles and the mass of barium peroxide, BaO2, needed to produce 1.5 kg of barium oxide, BaO (O2 is the other product.)
The Correct Answer and Explanation is:
To solve this problem, we need to follow these steps:
Step 1: Write the Balanced Chemical Equation
The reaction between barium peroxide (BaO2) and heat to form barium oxide (BaO) and oxygen (O2) is:2BaO2→2BaO+O22 \text{BaO}_2 \rightarrow 2 \text{BaO} + \text{O}_22BaO2→2BaO+O2
From the balanced equation, we see that 2 moles of BaO2 produce 2 moles of BaO and 1 mole of O2.
Step 2: Determine the Molar Mass of BaO
Next, we calculate the molar mass of BaO (barium oxide):
- The atomic mass of barium (Ba) is approximately 137.33 g/mol.
- The atomic mass of oxygen (O) is approximately 16.00 g/mol.
Thus, the molar mass of BaO is:137.33+16.00=153.33 g/mol137.33 + 16.00 = 153.33 \, \text{g/mol}137.33+16.00=153.33g/mol
Step 3: Convert 1.5 kg of BaO to Grams
Since the problem gives us 1.5 kg of BaO, we convert it to grams:1.5 kg=1500 g1.5 \, \text{kg} = 1500 \, \text{g}1.5kg=1500g
Step 4: Calculate the Moles of BaO Produced
Now, we calculate the number of moles of BaO in 1500 g:Moles of BaO=Mass of BaOMolar Mass of BaO=1500 g153.33 g/mol≈9.79 mol\text{Moles of BaO} = \frac{\text{Mass of BaO}}{\text{Molar Mass of BaO}} = \frac{1500 \, \text{g}}{153.33 \, \text{g/mol}} \approx 9.79 \, \text{mol}Moles of BaO=Molar Mass of BaOMass of BaO=153.33g/mol1500g≈9.79mol
Step 5: Use the Stoichiometric Relationship to Find the Moles of BaO2
From the balanced equation, we know that 2 moles of BaO2 produce 2 moles of BaO. Therefore, the moles of BaO2 required are the same as the moles of BaO produced. So, we need 9.79 moles of BaO2.
Step 6: Calculate the Mass of BaO2 Needed
The molar mass of BaO2 is:
- The atomic mass of barium (Ba) is 137.33 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol, and there are 2 oxygen atoms in BaO2, so the mass of oxygen in BaO2 is 32.00 g/mol.
Thus, the molar mass of BaO2 is:137.33+32.00=169.33 g/mol137.33 + 32.00 = 169.33 \, \text{g/mol}137.33+32.00=169.33g/mol
Now, we can calculate the mass of BaO2 needed:Mass of BaO2=Moles of BaO2×Molar Mass of BaO2=9.79 mol×169.33 g/mol≈1650 g\text{Mass of BaO2} = \text{Moles of BaO2} \times \text{Molar Mass of BaO2} = 9.79 \, \text{mol} \times 169.33 \, \text{g/mol} \approx 1650 \, \text{g}Mass of BaO2=Moles of BaO2×Molar Mass of BaO2=9.79mol×169.33g/mol≈1650g
Final Answer:
- Moles of BaO2 needed: 9.79 mol
- Mass of BaO2 needed: 1650 g or 1.65 kg
Thus, you would need approximately 9.79 moles or 1.65 kg of barium peroxide (BaO2) to produce 1.5 kg of barium oxide (BaO).
