Question

When 14.12 moles of mercury (ll) oxide decomposes, how many moles of oxygen gas will form ?

Answer 1

The number of moles of oxygen formed is 7.06 moles

Step-by-step explanation

Given that

The number of moles of mercury (II) oxide is 14.12 moles

Follow the steps below to find the number of moles of oxygen

Step 1; Write the balanced equation for the decomposition of the reaction

`\text{ 2HgO }\rightarrow\text{ 2Hg}_((s))\text{ + O}_(2(g))`

In the reaction above, 2 moles HgO decompose to produce 2 moles Hg and 1 mole O2

Step 2; Find the number of moles of oxygen using a stoichiometry ratio

Let x represents the number of moles of oxygen

`\begin{gathered} \text{ 2 moles HgO }\rightarrow\text{ 1 mole O}_2 \\ \text{ 14.12 moles HgO}\rightarrow\text{ x mole O}_2 \\ \text{ cross multiply} \\ \text{ 2 moles HgO }*\text{ x mole O}_2\text{ }=\text{ 1 mole O}_2*\text{ 14.12 moles HgO} \\ \text{ Isolate x} \\ \text{ x = }\frac{1\text{ mole O}_2*14.12moles\cancel{HgO}}{2moles\cancel{HgO}} \\ \text{ x = }(14.12)/(2) \\ \text{ x = 7.06 moles} \end{gathered}`

Therefore, the number of moles of oxygen formed is 7.06 moles