Question

Suppose 83.0 mL. of dioxygen gas are produced by this reaction, at a temperature of 70.0 °C and pressure of exactly I atm. Calculate the mass of mercury(II) oxide that must have reacted. Round your answer to 3 significant digits.

Answer 1

To find the mass of mercury(II) oxide that decomposed to produce 83.0 mL of dioxygen gas at 70.0 °C and 1 atm, use the ideal gas law to find the moles of O2, then use the stoichiometry of the HgO decomposition reaction, and finally calculate the mass using the molar mass of HgO.

Step-by-step explanation:

Calculating the Mass of Mercury(II) Oxide

To calculate the mass of mercury(II) oxide (HgO) that must have decomposed to produce 83.0 mL of dioxygen gas (O2), we can use the ideal gas law. First, we need to convert the volume of oxygen from milliliters to liters (83.0 mL = 0.0830 L). We then apply the ideal gas law PV = nRT, where P is the pressure (1 atm), V is the volume (0.0830 L), R is the ideal gas constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin (70.0 °C + 273.15 = 343.15 K).

First, solve for n (the number of moles of O2):
n = PV / RT = (1 atm)(0.0830 L) / (0.0821 L*atm/mol*K)(343.15 K)

This calculation gives us the moles of O2 produced. Next, we use the stoichiometry of the decomposition reaction of mercury(II) oxide, which is typically 2HgO → 2Hg + O2, to find the moles of HgO that decomposed. Since the stoichiometry shows a 1:1 molar ratio between HgO and O2, the moles of HgO will equal the moles of O2.Finally, we convert the moles of HgO to mass using the molar mass of HgO. The molar mass of HgO is approximately 216.59 g/mol.
Mass of HgO = moles of HgO * molar mass of HgO

Answer 2

m = 0.3249 g

Step-by-step explanation:

First, I'm assuming you have a reaction of mercury(II) oxyde descomposition. If this is the case, then the equation to use is the following:

HgO ---------> Hg + O2

Balancing the equation:

2HgO ----------> 2Hg + O2

This means that 2 moles of HgO reacts to produce 1 mole of O2, so, we first calculate the moles of O2, then, the moles of HgO and finally the mass:

We have the volume of O2, the pressure and temperature, so let's use the ideal gas equation:

PV = nRT

Solving for n:

n = PV/RT

R: 0.082 L atm / K mole

T = 70 + 273 = 343 K

V = 83 / 1000 = 0.083 L

Calculating n:

n = 1 * 0.083 / 0.082 * 343

n = 0.003 moles

as stated before, 2 moles of HgO reacts with 1 mole of oxygen so:

2 moles HgO = 1 moles O2

moles HgO = moles O2 / 2

moles HgO = 0.003 / 2 = 0.0015 moles

Finally, to calculate the mass:

m = n * MM

the molar mass of HgO is 216.59 g/mol, so replacing:

m = 0.0015 * 216.59

m = 0.3249 g