What is the maximum mass of the lower molar mass product that can be formed when 71.0 g of solid barium bromide and 44.5 g chlorine gas react?
BaBr2 + Cl2 ==> BaCl2 + Br2
This is a limiting reagent problem. I know that because BOTH reactants are given. I work these problems by solving two simple stoichiometry problems (simple meaning they are not limiting reagent problems and that usually gives you two answers for the product. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Here is a link that will show you how to do the stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
To determine the maximum mass of the lower molar mass product formed, we need to first determine the limiting reactant. The limiting reactant is the reactant that gets completely consumed in the reaction and determines the amount of product formed.
Let's start by writing the balanced chemical equation for the reaction between solid barium bromide (BaBr2) and chlorine gas (Cl2):
BaBr2 + Cl2 -> BaCl2 + Br2
From the balanced equation, we can see that it follows a 1:1 stoichiometric ratio between BaBr2 and Cl2, which means that one mole of BaBr2 reacts with one mole of Cl2 to produce one mole of BaCl2.
Next, we need to calculate the number of moles for each reactant using their molar masses. The molar mass of BaBr2 is the sum of the atomic masses of barium (Ba) and bromine (Br), which is 137.33 g/mol. The molar mass of Cl2 is simply the atomic mass of chlorine (Cl), which is 35.45 g/mol.
The number of moles for barium bromide (BaBr2) can be calculated by dividing the given mass (71.0 g) by its molar mass (137.33 g/mol):
Number of moles of BaBr2 = 71.0 g / 137.33 g/mol ≈ 0.5177 mol
Similarly, the number of moles for chlorine gas (Cl2) can be calculated by dividing the given mass (44.5 g) by its molar mass (35.45 g/mol):
Number of moles of Cl2 = 44.5 g / 35.45 g/mol ≈ 1.2559 mol
Now, we can compare the mole ratio between BaBr2 and Cl2 in the balanced equation. The ratio is 1:1, which means that 1 mole of BaBr2 reacts with 1 mole of Cl2.
Since we have more moles of Cl2 (1.2559 mol) compared to BaBr2 (0.5177 mol), the BaBr2 is the limiting reactant. This means that all 0.5177 moles of BaBr2 will react, and the Cl2 will be in excess.
To find the maximum mass of the lower molar mass product (BaCl2), we need to calculate it based on the limiting reactant (BaBr2) and its mole ratio to the product (BaCl2) in the balanced equation.
From the balanced equation, we see that the mole ratio between BaBr2 and BaCl2 is 1:1. This means that 1 mole of BaBr2 produces 1 mole of BaCl2.
So, the number of moles of BaCl2 formed will be equal to the number of moles of BaBr2 (0.5177 mol).
To find the maximum mass of BaCl2, we multiply the number of moles of BaCl2 by its molar mass. The molar mass of BaCl2 is the sum of the atomic masses of barium (Ba) and chlorine (Cl), which is 137.33 g/mol + 2 * 35.45 g/mol = 208.23 g/mol.
Maximum mass of BaCl2 = number of moles of BaCl2 * molar mass of BaCl2
Maximum mass of BaCl2 = 0.5177 mol * 208.23 g/mol ≈ 107.66 g
Therefore, the maximum mass of the lower molar mass product (BaCl2) that can be formed is approximately 107.66 grams.