B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)
How many moles of chlorine are used up in a reaction that produced 0.35 kg of BCl3
I do not understand please help
With some minor variations, all of these stoichiometry problems are alike.
Step 1. Write and balance the equation. You have that.
Step 2. Convert what you have to mols. That is mols = grams/molar mass = 350g/117 = approx 3 but you need to confirm all of these estimates here and those that follow.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (in this mols BCl3) to mols of what you want (in this case mols Cl2). That is done this way.
3 mols BCl3 x 3 mols Cl2/2 mol BCl3) = 3 x 3/2 = 9/2 = 4.5 mols Cl2. The problem stops here; i.e., the problem asks for mols Cl2.
Step 4. The usual question asks for grams Cl2. To find grams Cl2 it's g Cl2 = mols Cl2 x molar mass Cl2 = 4.5 x (2*35.44) = ?
Ask follow up question on what yu don't understand but explain in detail what you don't understand.
To find the number of moles of chlorine used up in the reaction, we can use the stoichiometry of the balanced chemical equation.
First, we need to determine the molar mass of BCl3.
The molar mass of boron (B) is approximately 10.81 g/mol, and the molar mass of chlorine (Cl) is approximately 35.45 g/mol. Since there are three chlorine atoms in each molecule of BCl3, the molar mass of BCl3 is:
Molar mass of BCl3 = (Mass of boron + 3 * Mass of chlorine) = (10.81 g/mol + 3 * 35.45 g/mol) = 137.17 g/mol
Now, we can convert the given mass of BCl3 (0.35 kg) to moles using the molar mass:
Number of moles of BCl3 = Mass of BCl3 / Molar mass of BCl3
= 0.35 kg * 1000 g/kg / 137.17 g/mol
≈ 2.5524 mol
From the balanced chemical equation, we can see that the stoichiometric ratio between BCl3 and Cl2 is 2:3. This means that for every 2 moles of BCl3 produced, we need 3 moles of Cl2.
Therefore, to find the number of moles of chlorine used up, we can use the ratio:
Number of moles of Cl2 = (Number of moles of BCl3) * (3 moles of Cl2 / 2 moles of BCl3)
= 2.5524 mol * (3/2)
≈ 3.8286 mol
So, approximately 3.83 moles of chlorine are used up in the reaction that produced 0.35 kg of BCl3.