A sample of mercury with a mass of 114.0 g was combined
with 12.8 g of oxygen gas, and the resulting reaction gave
123.1 g of mercury(II) oxide. How much oxygen was left
over after the reaction was complete?
The answer I got was 3.7 grams. Am I correct?
Well, it seems like you might have made a little mistake there. But don't worry, I'm here to clarify things for you! Let's do some calculations together and figure it out.
First, let's determine the total amount of oxygen that reacted with the mercury:
Mass of mercury = 114.0 g
Mass of oxygen = 12.8 g
Mass of mercury(II) oxide = 123.1 g
To find the mass of oxygen that reacted, we subtract the mass of mercury from the mass of mercury(II) oxide:
Mass of oxygen reacted = Mass of mercury(II) oxide - Mass of mercury
Mass of oxygen reacted = 123.1 g - 114.0 g
Now let's do the math:
Mass of oxygen reacted = 9.1 g
So, based on this calculation, 9.1 grams of oxygen reacted with the mercury. Therefore, the amount of oxygen left over after the reaction is complete would be:
Mass of oxygen left over = Mass of oxygen - Mass of oxygen reacted
Mass of oxygen left over = 12.8 g - 9.1 g
And finally:
Mass of oxygen left over = 3.7 g
Bingo! You got it right! So, congratulations on your correct answer of 3.7 grams. Great job!
To find the amount of oxygen gas that was left over after the reaction, we need to first determine the theoretical amount of mercury(II) oxide that would have been produced if all the mercury reacted with the oxygen.
Let's calculate the number of moles of mercury using its molar mass.
Molar mass of mercury (Hg) = 200.59 g/mol
Number of moles of Hg = Mass of Hg / Molar mass of Hg
Number of moles of Hg = 114.0 g / 200.59 g/mol
Number of moles of Hg ≈ 0.569 mol
The balanced chemical equation for the reaction is:
2Hg + O2 → 2HgO
From the balanced equation, we can see that the ratio of HgO to O2 is 2:1.
Therefore, the number of moles of HgO produced would be half of the number of moles of Hg, since the ratio is 2:1.
Number of moles of HgO produced = 0.569 mol / 2
Number of moles of HgO produced ≈ 0.285 mol
Now, let's calculate the molar mass of mercury(II) oxide (HgO).
Molar mass of HgO = 216.59 g/mol
Using the molar mass of HgO, we can calculate the mass of HgO produced.
Mass of HgO produced = Number of moles of HgO produced × Molar mass of HgO
Mass of HgO produced = 0.285 mol × 216.59 g/mol
Mass of HgO produced ≈ 61.76 g
To find the amount of oxygen gas left over, subtract the mass of HgO produced from the initial mass of oxygen gas.
Mass of oxygen gas left over = Mass of oxygen gas - Mass of HgO produced
Mass of oxygen gas left over = 12.8 g - 61.76 g
Mass of oxygen gas left over ≈ -48.96 g
The negative value indicates that there wasn't enough oxygen gas present to react completely with the mercury. Therefore, the answer is not 3.7 grams.
To solve this question, we need to use the concept of stoichiometry, which is the numerical relationship between the amounts of reactants and products in a chemical reaction.
First, we need to determine the balanced chemical equation for the reaction between mercury and oxygen gas. The balanced equation is:
2Hg + O2 → 2HgO
From the balanced equation, we can see that 2 moles of mercury react with 1 mole of oxygen gas to produce 2 moles of mercury(II) oxide.
To calculate the amount of oxygen reacted, we need to find the number of moles of mercury and oxygen gas involved in the reaction.
Given:
Mass of mercury = 114.0 g
Mass of oxygen gas = 12.8 g
Mass of mercury(II) oxide = 123.1 g
Step 1: Calculate the moles of mercury:
Molar mass of mercury (Hg) = 200.59 g/mol (from the periodic table)
Moles of mercury = Mass of mercury / Molar mass of mercury
Moles of mercury = 114.0 g / 200.59 g/mol ≈ 0.568 mol
Step 2: Calculate the moles of oxygen gas:
Molar mass of oxygen (O2) = 32.00 g/mol (from the periodic table)
Moles of oxygen gas = Mass of oxygen gas / Molar mass of oxygen
Moles of oxygen gas = 12.8 g / 32.00 g/mol = 0.40 mol
Step 3: Determine the stoichiometry of mercury to oxygen gas:
From the balanced equation, we know that 2 moles of mercury react with 1 mole of oxygen gas. Therefore, the ratio of mercury to oxygen gas is 2:1.
Step 4: Calculate the moles of oxygen gas required for the reaction:
Moles of oxygen gas required = Moles of mercury / Ratio of mercury to oxygen gas
Moles of oxygen gas required = 0.568 mol / 2 ≈ 0.284 mol
Step 5: Calculate the moles of oxygen gas remaining:
Moles of oxygen gas remaining = Moles of oxygen gas initially - Moles of oxygen gas required
Moles of oxygen gas remaining = 0.40 mol - 0.284 mol ≈ 0.116 mol
Finally, convert the moles of oxygen gas remaining to grams:
Mass of oxygen gas remaining = Moles of oxygen gas remaining * Molar mass of oxygen gas
Mass of oxygen gas remaining = 0.116 mol * 32.00 g/mol ≈ 3.712 g
Therefore, the amount of oxygen gas left over after the reaction is approximately 3.712 grams. Thus, your answer of 3.7 grams is correct.
114.0g Hg = 0.568 moles
12.8g O2 = 0.400 moles = 0.8 moles of O
123.1g of HgO = 0.568 moles HgO
so, 0.232 moles of O = 3.712g
Looks good to me