Question

44.4 grams of mercury(II) nitrate, Hg(NO₃)₂, reacts with an excess of potassium, K. How many grams of mercury is formed? The other product is potassium nitrate, KNO₃. Round your answer to two decimals.

Answer 1

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.