Mercury(II) oxide ([tex] HgO [/tex]) decomposes to form mercury ([tex] Hg [/tex]) and oxygen ([tex] O_2 [/tex]). The balanced chemical equation is shown below:

[tex]\[ 2 HgO \rightarrow 2 Hg + O_2 \][/tex]

The molar mass of [tex] O_2 [/tex] is [tex] 32.00 \, \text{g/mol} [/tex]. How many moles of [tex] HgO [/tex] are needed to produce [tex] 250.0 \, \text{g} [/tex] of [tex] O_2 [/tex]?

A. 3.906 moles

B. 7.813 moles

C. 15.63 moles

D. 73.87 moles



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]