Consider the reaction
2D2(g)+O2(g)→2D2O(l)
What is the mass of heavy water, D2O(l), produced when 3.75g of O2(g) reacts with excess D2(g)?
Well, let's grab our calculators and dive into the problem!
To find the mass of D2O(l) produced, we first need to determine the number of moles of O2(g) that reacted. We can do that by using the molar mass of O2, which is about 32 g/mol.
So, 3.75g of O2 is equal to 3.75g divided by 32 g/mol, which is approximately 0.1172 mol.
Since the reaction is 2D2(g) + O2(g) → 2D2O(l), we see that every 1 mole of O2 reacts with 2 moles of D2O.
So, if we have 0.1172 mol of O2, it will react with twice as many moles of D2O. That means we will have 2 * 0.1172 mol of D2O.
So, the number of moles of D2O produced is approximately 0.2344 mol.
Now, to find the mass of D2O, we can use its molar mass, which is about 20 g/mol. Multiplying that by the number of moles, we get:
0.2344 mol * 20 g/mol = 4.688 g
Therefore, the mass of heavy water, D2O(l), produced when 3.75g of O2(g) reacts with excess D2(g) is approximately 4.688 grams.
Keep in mind that this is just a rough estimate, so don't take it too seriously!
To determine the mass of heavy water, D2O(l), produced in the reaction, we need to use stoichiometry.
The balanced equation for the reaction is:
2D2(g) + O2(g) → 2D2O(l)
From the equation, we can see that the stoichiometric ratio between O2 and D2O is 1:2. This means that for every 1 mole of O2, we will obtain 2 moles of D2O.
Step 1: Convert the given mass of O2 to moles.
Molar mass of O2 = 32 g/mol
Number of moles of O2 = mass / molar mass
= 3.75 g / 32 g/mol
≈ 0.1172 mol
Step 2: Use the stoichiometric ratio to find the moles of D2O produced.
According to the stoichiometry, 1 mole of O2 produces 2 moles of D2O.
Number of moles of D2O = 2 moles of D2O / 1 mole of O2 * 0.1172 mol of O2
= 0.1172 mol * 2
= 0.2344 mol
Step 3: Convert the moles of D2O to mass.
Molar mass of D2O = 20 g/mol
Mass of D2O = moles of D2O * molar mass
= 0.2344 mol * 20 g/mol
= 4.688 g
Therefore, the mass of heavy water, D2O(l), produced when 3.75 g of O2(g) reacts with excess D2(g), is approximately 4.688 g.
To find the mass of heavy water, D2O(l), produced in this reaction, we need to use stoichiometry.
First, let's analyze the given reaction:
2D2(g) + O2(g) → 2D2O(l)
From the balanced equation, we can see that it takes 1 mole of O2 to produce 2 moles of D2O.
To find the moles of O2 used in the reaction, we can use the following formula:
moles = mass / molar mass
The molar mass of O2 is 32.00 g/mol. Therefore, the moles of O2 can be calculated as follows:
moles of O2 = 3.75g / 32.00 g/mol = 0.1172 mol
Now that we know the moles of O2 used, we can determine the moles of D2O produced.
moles of D2O = (moles of O2) x (moles of D2O / moles of O2)
From the balanced equation, we can see that the molar ratio of O2 to D2O is 1:2.
moles of D2O = 0.1172 mol x (2 mol D2O / 1 mol O2) = 0.2344 mol
Finally, we can calculate the mass of D2O produced using the formula:
mass = moles x molar mass
The molar mass of D2O is 20.03 g/mol. Therefore, the mass of D2O produced can be calculated as follows:
mass of D2O = 0.2344 mol x 20.03 g/mol = 4.692 g
Therefore, the mass of heavy water, D2O(l), produced when 3.75g of O2(g) reacts with excess D2(g), is approximately 4.692 g.
mols O2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols O2 to mols D2O.
Now convert mols D2O to grams. g = mols x molar mass = ?