Questions: Suppose 11.0 mL of dioxygen gas are produced by this reaction, at a temperature of 110.0°C and pressure of exactly 1 atm. Calculate the mass of mercury(II) oxide that must have reacted. Round your answer to 3 significant digits. g

Suppose 11.0 mL of dioxygen gas are produced by this reaction, at a temperature of 110.0°C and pressure of exactly 1 atm. Calculate the mass of mercury(II) oxide that must have reacted. Round your answer to 3 significant digits.
g

Transcript text: 2. Suppose 11.0 mL of dioxygen gas are produced by this reaction, at a temperature of $110.0^{\circ} \mathrm{C}$ and pressure of exactly 1 atm . Calculate the mass of mercury(II) oxide that must have reacted. Round your answer to 3 significant digits. $\square$ g

Solution

Solution Steps

Step 1: Convert the volume of O2 gas produced to moles

Using the ideal gas law, $PV = nRT$, where $P = 1$ atm, $V = 11$ L, $R = 0.0821$ L·atm/(mol·K), and $T = 383.15$ K, we find the moles of $\mathrm{O}_2$ gas produced to be $n = 0.35$ moles.

Step 2: Calculate the moles of HgO reacted

Based on the stoichiometry of the balanced chemical equation, $2 \mathrm{HgO}(s) \rightarrow 2 \mathrm{Hg}(l) + \mathrm{O}_{2}(g)$, the moles of $\mathrm{HgO}$ reacted is twice the moles of $\mathrm{O}_2$ produced, which gives $n = 0.699$ moles of $\mathrm{HgO}$.

Step 3: Convert moles of HgO to mass

Multiplying the moles of $\mathrm{HgO}$ by its molar mass ($216.59$ g/mol), we find the mass of $\mathrm{HgO}$ that must have reacted to be $0.699\times 216.59 = 151.478 g$.