For the First One, I believe you can choose any one since they both have equal a
ID: 713800 • Letter: F
Question
For the First One, I believe you can choose any one since they both have equal amount of initial reactants and the same number of coefficients. Now I know the answer to number 1. My issue is why is it that you have to multiply by 2 moles to get the answer.You’re going to get 1.5 moles for either one reactant.
So for MnO2: 1.5mol *86.94g/mol = 130.41 g of Mn O2. Now the answer is 130.41 * 2= 260.82
My Question is why can’t you double the mass number when doing the conversion.
Which is 1.5 *173.88g/2Mol. It’s a different answer why? Please answer my question. Do not just put up the solution for Problem1. Problem 3 is a little different. I have to convert grams to moles then do reactant product ratio. But I keep getting a high number. The answer is 8.18g. Please help
3. For the reaction shown, compute the theoretical yield of the product (in grams) for each of the following initial amounts of reactants: Ti (s) + 2 F2 (g) TiF4 (s) a. 5.0 g Ti, 5.0 g F2 ANS: 8.18 g
Explanation / Answer
1) Initial amount of Mn=3mol
initial amount of O2=3mol
Whatever be the amount of reactants ,you must only be concerned about the reaction stoichiometry with respect to product that is:
2 mol of Mn produces 2 mol MnO2 ,molar ratio of Mn:MnO2=1:1
and that 2mol of O2 produces 2mol MnO2 ,molar ratio of O2:MnO2=1:1
Starting with O2 as the reactant , 3mol of O2 produces (3mol of MnO2)
and
Starting with Mn as the reactant , 3mol of Mn produces (3mol of MnO2)
In both cases,theoretical yield of MnO2=(3mol of MnO2)=3mol *86.94g/mol=260.82 g
3)Initial mol of Ti=mass of Ti/molar mass of Ti=5.0g/47.867g/mol=0.104mol
Initial mol of F2=mass of F2/molar mass of F2=5.0g/37.997 g/mol=0.131mol
Reaction stoichiometry:
molar ratio of Ti : F2=1:2
So, 0.104mol of Ti would require 2*0.104 mol (=0.208 mol)of F2 to completely react
As the given mol of F2 is less that ,what Ti requires, so Ti is the excess reactant and F2 is the limiting reactant .
The limiting reactant is completely used up to form the product ,while someamount of the excess reactant is left over unreacted.
So proceeding with 0.131mol F2, molar ratio of F2:TiF4=2:1
Thus ,mol of TiF4 produced=2*mol of F2=1/2*0.131mol=0.0655 mol
mass of TiF4 producd=0.0655mol*molar mass of TiF4=0.0655mol*123.861g/mol=8.112g