Consider the reaction
2D2(g)+O2(g)→2D2O(l)
What is the mass of heavy water, D2O(l), produced when 7.65g of O2(g) reacts with excess D2(g)?
Express your answer with the appropriate units.
Well, if you ask me, this reaction sounds like a real "heavyweight" bout in the chemical world! Let's break it down.
According to the balanced equation, we see that the molar ratio between O2 and D2O is 1:2.
To find the mass of D2O produced, we first need to find the number of moles of O2 reacted. We can do that by dividing the given mass of O2 by its molar mass, which is about 32 g/mol. So, 7.65 g / 32 g/mol gives us the number of moles of O2.
Now, using the molar ratio from the balanced equation, we can determine the number of moles of D2O that will be produced. Since the ratio is 1:2 (O2 to D2O), we can multiply the number of moles of O2 by 2 to get the number of moles of D2O.
To determine the mass of D2O, we just multiply the number of moles of D2O by its molar mass, which is about 20 g/mol.
So, in short, we take the number of moles of O2 and multiply it by 2 to get the number of moles of D2O, and then multiply that by the molar mass of D2O to get the mass.
But since you've already done half the work and neglected the D2, let me just cut to the chase. The mass of D2O produced would be zero because there's no D2 in the equation to react with the O2. Looks like it's a no-go for the heavy water this time!
To find the mass of heavy water, D2O(l), produced, we need to use the given reaction equation and stoichiometry.
The balanced reaction equation is:
2D2(g) + O2(g) → 2D2O(l)
From the equation, we can see that 1 mole of O2 reacts with 2 moles of D2 to produce 2 moles of D2O.
First, we need to determine the number of moles of O2 present using its given mass and its molar mass. The molar mass of O2 is 32.00 g/mol (16.00 g/mol for each oxygen atom).
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 7.65 g / 32.00 g/mol = 0.239 mole
Since the reaction is balanced, we know that the number of moles of D2O produced is equal to the number of moles of O2 used in the reaction.
Moles of D2O = Moles of O2 = 0.239 mole
Now, to find the mass of D2O produced, we need to multiply the moles of D2O by its molar mass. The molar mass of D2O is 20.03 g/mol (because 2(2.01 g/mol) + 16.00 g/mol = 20.03 g/mol).
Mass of D2O = Moles of D2O × Molar mass of D2O
Mass of D2O = 0.239 mole × 20.03 g/mol = 4.79 g
Therefore, the mass of heavy water, D2O(l), produced when 7.65 g of O2(g) reacts with excess D2(g) is 4.79 g.