How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following chemical equation? MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
Answers
The answer should be 14.6g.
30g HCl = .823mol
.823mol HCl => .206mol Cl2 (Cl2 is produced in a 1:4 ratio)
.206mol Cl2 = 14.6g
If the question has "percent yield" in it somewhere, then if your ideal yield is only 89% your maximum yield will be 13g.
To check your answer, here's a stoichiometry calculator I made back when I was a sophomore.
Reaction Stoichiometry Calculator
Note: You can never go wrong with tried and true grams to moles, moles to grams, but Debarshi's method is a little easier.
Answer:
The answer is 13.0g Cl2, or 13.048g Cl2 to be exact.
Explanation:
When finding the mass of a specific chemical formula in a balanced equation, it is crucial to find the limiting reagent first, that way we are able to identify the BEST possible answer.
Sure, if we solved for both the 30.0g of HCl and the 16.0g of MnO2, we would get two answers: 13.048 (produced from 16.0g of MnO2) and 14.6 (produced from 30.0g of HCl). However, there can only be one answer. And we find this by determining the limiting reagent.
When solving for the limiting reagent, it is easiest to remember the rule of three, or x=(c)(b)/a (or basic conversions)
Let's start with grams of MnO2. To do this, we take the opposite measurement of the chemical formula that is on the same side of the balanced equation, or HCl. In other words, 30.0g HCl.
When dealing with conversions, it easier to breakdown the formulas we need and cancel out the ones we don't.
So, we'll be multiplying 30.0g HCl by:
x/the molar mass of HCl (the molar mass serves as the grams of HCl, or in this case 36.458g HCl)
With this, HCl cancels out. But, we need grams of MnO2. Easy. We'll put the molar mass of MnO2 on top of the molar mass of HCl. That gives us:
(30.0gHCl)(86.93684gMnO2) / (36.458gHCl)
Notice how the calculation above holds the same structure as our rule of three: x=(c)(b)/a
When we put that in a calculator, we get 71.5372538g MnO2
However, we only have 16.0g of MnO2. So this is our limiting reagent.
Now that we have our limiting reagent, we can find our grams of Cl2.
To do this, we do the same thing as before, but now we use 16.0g MnO2 as our multiplier.
Like before, we must cancel out grams of MnO2, because we're trying to find Cl2. So, substitute the molar mass of MN02 on the bottom.
Now, add the molar mass of Cl2 to the top.
This should get us:
Grams of Cl2 -> (16.0g MnO2) x (70.9g Cl2) / (86.93684g MnO2)
When we put all of that into a calculator, we get 13.0g Cl2.