Final answer:
When 44.4 grams of mercury(II) nitrate reacts with potassium, 27.42 grams of mercury will be formed after the reaction.
Step-by-step explanation:
To determine how many grams of mercury is formed when 44.4 grams of mercury(II) nitrate reacts with an excess of potassium, we must first write a balanced chemical equation for the reaction:
Hg(NO3)2 + 2 K → Hg + 2 KNO3
Next, we'll calculate the molar mass of mercury(II) nitrate, which is (200.59 g/mol for Hg) + (2 × 14.01 g/mol for N) + (2 × (3 × 16.00 g/mol for O)) = 200.59 + 28.02 + 96 = 324.61 g/mol.
Using this molar mass, we convert the mass of mercury(II) nitrate to moles:
44.4 g ÷ 324.61 g/mol = 0.1367 moles of Hg(NO3)2
Since the reaction shows that one mole of Hg(NO3)2 produces one mole of Hg, we'll have the same number of moles of mercury:
0.1367 moles of Hg × 200.59 g/mol for Hg = 27.42 grams of Hg
Therefore, 27.42 grams of mercury will be formed, rounded to two decimals.