Answer:
7.88 g
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.
[tex]M_{r}[/tex]: 200.59 32.00 216.59
2Hg + O₂ ⟶ 2HgO
Mass/g: 7.00 7.00
Step 2. Calculate the moles of each reactant
[tex]\text{Moles of Hg} = \text{7.00 g Hg} \times \frac{\text{1 mol Hg}}{\text{200.59 g Hg}} = \text{0.034 90 mol Hg}[/tex]
[tex]\text{Moles of O}_{2} = \text{7.00 g O}_{2} \times \frac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.2188 mol O}_{2}[/tex]
Step 3. Identify the limiting reactant
Calculate the moles of HgO we can obtain from each reactant.
From Hg: [tex]\text{Moles of HgO} = \text{0.034 90 mol Hg} \times \frac{\text{2 mol HgO} }{\text{2 mol Hg} } = \text{0.034 90 mol HgO}[/tex]
From O₂: [tex]\text{Moles of HgO} = \text{0.2188 mol O}_{2} \times \frac{\text{2 mol HgO} }{\text{1 mol O}_{2} } = \text{0.4375 mol HgO}[/tex]
The limiting reactant is Hg because it gives the smaller amount of HgO.
Step 4. Calculate the mass of HgO that you can obtain from Hg.
[tex]\text{Mass of HgO} = \text{0.034 90mol HgO} \times \frac{\text{216.59 g HgO}}{\text{1 mol HgO}} = \text{7.56 g HgO}[/tex]
