0.300 moles of bromine gas (Br2) and 0.500 moles of chlorine gas (Cl2) react to for tribromochlorine (Br3Cl). how many moles of this product will form ?
To determine how many moles of tribromochlorine (Br3Cl) will form, we can use the concept of stoichiometry. The balanced chemical equation for the reaction is:
Br2 + Cl2 → Br3Cl
From the equation, we can see that one mole of Br2 reacts with one mole of Cl2 to produce one mole of Br3Cl.
Given the amounts of reactants, we have 0.300 moles of Br2 and 0.500 moles of Cl2. Since the reaction occurs in a 1:1 ratio, we can deduce that the limiting reactant is Cl2, as it is in excess. This means that all of the Cl2 will be consumed in the reaction, and the amount of product formed will be determined by the amount of Br2.
Therefore, the number of moles of Br3Cl produced will also be 0.300 moles.
So, 0.300 moles of tribromochlorine (Br3Cl) will form.
A LR problem. See you third post up the list. Also, here is a site that gives specific instructions how to solve a limiting reagent problem.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html