Consider the following reaction:
C6H5CL+C2HOCL3=>C14H9CL5+H20
If 1142 g of C6H5CL is reacted with 485 g of C2HOCL3, what mass of C14H9CL5 would be formed?
Since an amount is given for both reactants we know this is a limiting reagent problem. I solve these by solving stoichiometry problems. First I choose one reactant, ignore the other one, and calculate the moles of product formed. Then I choose the the other reactant, ignore the other one, and calculate the moles of product formed. The answers will be different, of course, and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent providing that value is the limiting reagent. So how do you do all of this. Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of C14H9Cl5 formed in the reaction, we need to use stoichiometry and the given masses of C6H5Cl and C2HOCl3.
Here's how to approach the problem:
1. Write a balanced chemical equation for the reaction:
C6H5Cl + C2HOCl3 → C14H9Cl5 + H2O
2. Determine the molar masses of each compound:
- C6H5Cl: 12.01 g/mol (C) + 1.01 g/mol (H) + 35.45 g/mol (Cl) = 112.57 g/mol
- C2HOCl3: 12.01 g/mol (C) + 1.01 g/mol (H) + 35.45 g/mol (Cl) + 3 × 16.00 g/mol (O) = 163.33 g/mol
- C14H9Cl5: 12.01 g/mol (C) + 1.01 g/mol (H) + 35.45 g/mol (Cl) = 250.63 g/mol
3. Convert the given masses of C6H5Cl and C2HOCl3 to moles:
- Moles of C6H5Cl = 1142 g / 112.57 g/mol = 10.14 mol
- Moles of C2HOCl3 = 485 g / 163.33 g/mol = 2.97 mol
4. Use the stoichiometry of the balanced equation to determine the moles of C14H9Cl5 produced:
From the balanced equation, the molar ratio between C6H5Cl and C14H9Cl5 is 1:1. Therefore, the moles of C14H9Cl5 produced will be the same as the moles of C6H5Cl used.
5. Convert the moles of C14H9Cl5 to mass:
- Mass of C14H9Cl5 = Moles of C14H9Cl5 × Molar mass of C14H9Cl5
- Mass of C14H9Cl5 = 10.14 mol × 250.63 g/mol = 2547 g
Therefore, the mass of C14H9Cl5 formed in the reaction is approximately 2547 grams.