Tungsten(VI) oxide can be reduced to tungsten by heating it with carbo, as shown in the reaction:
2WO3(s) + 3C(s) <---> 2W(s) + 3CO2(g)
If the original mass of WO3 is 37.0 g, how many grams of CO2 must be collected from the reaction in order to infer that the solid that remains is pure tungsten? (Round to the nearest whole number)
I'm not quite understanding if I should take the molar mass of WO3 and CO2 and subtract them or if I should get the number of moles for each and find the mass that way.
I don't think you have the right picture.
Do it this way.
mols WO3 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols WO3 to mols CO2.
Now convert mols CO2 to grams CO2. g CO2 = mols CO2 x molar mass CO2. This gives you the grams of CO2 needed to obtain a 100% conversion of WO3 to W, then round to the nearest whole number.
Ah, I see you're getting down to the nitty-gritty of stoichiometry! Let's crunch those numbers and find out how much CO2 we need.
According to the balanced equation, 2 moles of WO3 reacts with 3 moles of C to produce 3 moles of CO2.
To find the moles of WO3 present, we'll use its molar mass. Tungsten has a molar mass of 183.84 g/mol, and oxygen has a molar mass of 16.00 g/mol. So, the molar mass of WO3 is 183.84 g/mol + (3 × 16.00 g/mol) = 231.84 g/mol.
Using the given mass of 37.0 g, we can calculate the moles of WO3 as follows:
Moles of WO3 = Mass of WO3 / Molar mass of WO3
= 37.0 g / 231.84 g/mol
≈ 0.1596 mol
Now, let's use the mole ratio from the balanced equation to find the moles of CO2 that would be produced:
Moles of CO2 = (3 mol CO2 / 2 mol WO3) × Moles of WO3
= (3 / 2) × 0.1596 mol
≈ 0.2394 mol
Finally, we can convert this moles of CO2 to grams:
Mass of CO2 = Moles of CO2 × Molar mass of CO2
= 0.2394 mol × (12.01 g/mol + 2 × 16.00 g/mol)
≈ 8.27 g
So, to infer that the solid remaining is pure tungsten, we would need to collect approximately 8 grams of CO2. Keep in mind that this value should be rounded to the nearest whole number as per your request.
Hope this helps! It's always fun to see chemicals clashing in balanced equations.
To solve this problem, we need to determine the stoichiometry of the reaction, which relates the number of moles of each reactant and product.
The balanced equation shows that 2 moles of WO3 react with 3 moles of C to produce 2 moles of W and 3 moles of CO2.
To find the number of moles of CO2 produced, we first need to calculate the moles of WO3.
1. Calculate the molar mass of WO3:
W molar mass = 183.84 g/mol
O molar mass = 16.00 g/mol (x 3 since there are three oxygen atoms in each WO3 molecule) = 48.00 g/mol
Molar mass of WO3 = 183.84 g/mol + 48.00 g/mol = 231.84 g/mol
2. Calculate the number of moles of WO3:
Moles = Mass / Molar mass = 37.0 g / 231.84 g/mol = 0.1593 mol
3. Using the stoichiometry of the balanced equation, we can determine the number of moles of CO2 produced:
Since the mole ratio between WO3 and CO2 is 2:3,
Moles of CO2 produced = (3/2) x Moles of WO3 = (3/2) x 0.1593 mol ≈ 0.239 mol
Now, to find the mass of CO2 produced, we can use its molar mass:
4. Calculate the molar mass of CO2:
C molar mass = 12.01 g/mol
O molar mass = 16.00 g/mol (x 2 since there are two oxygen atoms in each CO2 molecule) = 32.00 g/mol
Molar mass of CO2 = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
5. Calculate the mass of CO2 produced:
Mass = Moles x Molar mass = 0.239 mol x 44.01 g/mol ≈ 10.53 g
Rounding to the nearest whole number, we find that approximately 11 grams of CO2 must be collected from the reaction to infer that the solid that remains is pure tungsten.