Final answer:
To determine the mass of lead(II) chloride produced, we first identify the limiting reactant based on stoichiometry. In this reaction, Pb(NO3)2 is the limiting reactant. The mass of PbCl2 produced is calculated to be 2.26 grams.
Step-by-step explanation:
The student is asking how to calculate the mass of lead(II) chloride produced in a reaction between RbCl and Pb(NO3)2. To solve this, we need to consider stoichiometry and the concept of limiting reactant.
First, calculate the number of moles of RbCl using the provided volume and molarity:
Number of moles of RbCl = 0.025 L × 0.562 M = 0.01405 moles.
Next, calculate the moles of Pb(NO3)2:
Number of moles of Pb(NO3)2 = 0.020 L × 0.407 M = 0.00814 moles.
The stoichiometric ratio between Pb(NO3)2 and PbCl2 is 1:1. Since Pb(NO3)2 has fewer moles, it's the limiting reactant. Thus, the moles of PbCl2 produced will also be 0.00814 moles.
To find the mass, we multiply the moles by the molar mass of PbCl2 (which is 278.1 g/mol):
Mass of PbCl2 = 0.00814 moles × 278.1 g/mol = 2.26 g.
Therefore, the mass of lead(II) chloride produced in the reaction is 2.26 grams.