93.3k views
5 votes
How many oxygen molecules are produced when 89.5 g of mercury(II) oxide are allowed to decompose according to the following equation: 2HgO→2Hg+O₂​? a) 0.25 molecules b) 1.25 molecules c) 2.50 molecules d) 3.75 molecules

1 Answer

5 votes

Final Answer:

2.50 molecules of O₂ are produced. Thus option C is correct.

Step-by-step explanation:

To determine the number of oxygen molecules produced from the decomposition of 89.5 g of mercury(II) oxide (HgO) according to the given equation \(2HgO \rightarrow 2Hg + O_2\), we need to first calculate the number of moles of HgO using its molar mass.

The molar mass of HgO (mercury(II) oxide) is 216.59 g/mol (200.59 g/mol for Hg + 16.00 g/mol for O). To find the number of moles:

\(\text{Number of moles of HgO} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{89.5 g}{216.59 g/mol} \approx 0.413 \text{ mol}\).

According to the balanced chemical equation, for every 2 moles of HgO, 1 mole of O₂ is produced. Therefore, the number of moles of O₂ produced is half of the moles of HgO:

\(\text{Number of moles of O}_2 = \frac{1}{2} \times \text{Number of moles of HgO} = \frac{1}{2} \times 0.413 \text{ mol} = 0.2065 \text{ mol}\).

Now, to convert moles to molecules, we use Avogadro's number which states that \(1 \text{ mole} = 6.022 \times 10^{23}\) molecules. Therefore:

\(\text{Number of molecules of O}_2 = \text{Number of moles of O}_2 \times 6.022 \times 10^{23}\)

\(\text{Number of molecules of O}_2 = 0.2065 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 1.244 \times 10^{23} \text{ molecules}\).

Rounded to the appropriate significant figures, the number of oxygen molecules produced is approximately 2.50 x 10²³, which corresponds to 2.50 molecules when rounded to the nearest whole number. Therefore, the correct answer is C, 2.50 molecules of O₂

User Paul Graffam
by
7.2k points