Answer :
To solve this problem, let's break it down step-by-step:
1. Determine the molar mass of \(O_2\):
The molar mass of \(O_2\) is given as \(32.00 \; \text{g/mol}\).
2. Calculate the number of moles of \(O_2\) produced:
The mass of \(O_2\) produced is given as \(250.0 \; \text{g}\). To find the number of moles of \(O_2\), use the formula:
[tex]\[ \text{moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of } O_2 = \frac{250.0 \; \text{g}}{32.00 \; \text{g/mol}} = 7.8125 \; \text{moles} \][/tex]
3. Use the stoichiometry of the balanced chemical equation:
The balanced equation is:
[tex]\[ 2 \; HgO \rightarrow 2 \; Hg + O_2 \][/tex]
This tells us that 2 moles of \(HgO\) produce 1 mole of \(O_2\). Therefore, if we need to produce \(7.8125\) moles of \(O_2\), we can set up the proportion:
[tex]\[ \text{moles of } HgO = 2 \times \text{moles of } O_2 \][/tex]
Substituting the moles of \(O_2\):
[tex]\[ \text{moles of } HgO = 2 \times 7.8125 = 15.625 \; \text{moles} \][/tex]
4. Conclusion:
Therefore, to produce \(250.0 \; \text{g}\) of \(O_2\), \(15.625\) moles of \(HgO\) are needed.
Among the given choices, the closest value is \(15.63 \; \text{moles}\).
The answer is:
[tex]\[ 15.63 \; \text{moles} \][/tex]
1. Determine the molar mass of \(O_2\):
The molar mass of \(O_2\) is given as \(32.00 \; \text{g/mol}\).
2. Calculate the number of moles of \(O_2\) produced:
The mass of \(O_2\) produced is given as \(250.0 \; \text{g}\). To find the number of moles of \(O_2\), use the formula:
[tex]\[ \text{moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of } O_2 = \frac{250.0 \; \text{g}}{32.00 \; \text{g/mol}} = 7.8125 \; \text{moles} \][/tex]
3. Use the stoichiometry of the balanced chemical equation:
The balanced equation is:
[tex]\[ 2 \; HgO \rightarrow 2 \; Hg + O_2 \][/tex]
This tells us that 2 moles of \(HgO\) produce 1 mole of \(O_2\). Therefore, if we need to produce \(7.8125\) moles of \(O_2\), we can set up the proportion:
[tex]\[ \text{moles of } HgO = 2 \times \text{moles of } O_2 \][/tex]
Substituting the moles of \(O_2\):
[tex]\[ \text{moles of } HgO = 2 \times 7.8125 = 15.625 \; \text{moles} \][/tex]
4. Conclusion:
Therefore, to produce \(250.0 \; \text{g}\) of \(O_2\), \(15.625\) moles of \(HgO\) are needed.
Among the given choices, the closest value is \(15.63 \; \text{moles}\).
The answer is:
[tex]\[ 15.63 \; \text{moles} \][/tex]
