For the reaction:
V2O5 + 6HCl --> 2VOCl3 + 3H2O
If 13.65g of VOCl3 was produced in this reaction, how many grams of V2O5 were reacted in excess HCl?
I used stoichiometry and converted 13.65g of VOCl3 into grams of V2O5. I got 73.28 g. However, I feel like this isn't right because I didn't incorporate the excess HCl?
Thanks for your help.
To determine the number of grams of V2O5 that were reacted in excess, you need to consider the stoichiometry of the reaction and the molar ratios involved.
Let's start by examining the balanced chemical equation:
V2O5 + 6HCl --> 2VOCl3 + 3H2O
From the equation, we can see that the molar ratio between V2O5 and VOCl3 is 1:2. This means that for every 1 mole of V2O5 reacted, 2 moles of VOCl3 are produced.
To determine the number of moles of VOCl3 produced, we can use the given mass of 13.65g of VOCl3. To convert grams to moles, we need to know the molar mass of VOCl3, which is calculated by adding the atomic masses of vanadium (V), oxygen (O), and chlorine (Cl).
The molar mass of V: 51.0 g/mol
The molar mass of O: 16.0 g/mol (there are 3 oxygen atoms in VOCl3)
The molar mass of Cl: 35.5 g/mol (there are 3 chlorine atoms in VOCl3)
So the molar mass of VOCl3 = (51.0 g/mol) + (3 * 16.0 g/mol) + (3 * 35.5 g/mol) = 173.5 g/mol
To convert grams of VOCl3 to moles, we divide the mass (13.65g) by the molar mass (173.5 g/mol):
Number of moles of VOCl3 = 13.65g / 173.5 g/mol
Now that we have the number of moles of VOCl3 produced, we can use the stoichiometry of the equation to determine the number of moles of V2O5 reacted. According to the balanced equation, the molar ratio between V2O5 and VOCl3 is 1:2. This means that for every 1 mole of V2O5 reacted, 2 moles of VOCl3 are produced.
So, if the number of moles of VOCl3 is X, then the number of moles of V2O5 is X/2.
Now, we can convert the number of moles of V2O5 to grams by multiplying by its molar mass (181.9 g/mol) to find the grams of V2O5 reacted in excess HCl:
Grams of V2O5 reacted in excess HCl = (X/2) * molar mass of V2O5
As you can see, we need to know the number of moles of VOCl3 produced (X) before we can calculate the grams of V2O5. In this case, we have the grams of VOCl3 (13.65g), but we need to convert it to moles in order to proceed:
Number of moles of VOCl3 = 13.65g / molar mass of VOCl3
By using these calculations, you can determine the grams of V2O5 that were reacted in excess HCl.