Question

40.0 g of BaCl₂ is added to 1.0 L of 0.50 M of AgNO₃. The following reaction occurs:

2AgNO₃(aq)+BaCl₂(aq)→2AgCl(s)+Ba(NO₃)₂(aq)
Calculate the mass of the precipitate, AgCl, that will be produced and the moles of the reactant in excess that will remain in the solution after the reaction is complete.

Answer 1

The mass of the precipitate, AgCl, that will be produced is 55.026 g. The moles of the reactant in excess that will remain in the solution after the reaction is complete is 0.116 mol.

Step-by-step explanation:

To calculate the mass of the precipitate, AgCl, that will be produced, we need to determine the limiting reactant. In this case, we have 40.0 g of BaCl₂ and 1.0 L of 0.50 M AgNO₃. We can use the stoichiometry of the balanced equation to find the moles of AgCl produced, and then convert to grams.

First, let's calculate the moles of BaCl₂ using its molar mass (BaCl₂: 208.23 g/mol). Moles of BaCl₂ = (mass of BaCl₂) / (molar mass of BaCl₂) = (40.0 g) / (208.23 g/mol) = 0.192 mol BaCl₂

Next, we can use the coefficients of the balanced equation to relate the moles of BaCl₂ to the moles of AgCl. From the balanced equation, we can see that 2 moles of AgCl are produced for every 1 mole of BaCl₂ consumed. Therefore, moles of AgCl = (2 moles AgCl / 1 mole BaCl₂) * (0.192 mol BaCl₂) = 0.384 mol AgCl

Finally, we can convert the moles of AgCl to grams using its molar mass (AgCl: 143.32 g/mol). Mass of AgCl = (moles of AgCl) * (molar mass of AgCl) = (0.384 mol) * (143.32 g/mol) = 55.026 g AgCl

To find the moles of the reactant in excess that will remain in the solution, we can calculate the moles of AgNO₃ used in the reaction. We know the volume and concentration of the AgNO₃ solution, so we can again use the equation C = n / V to solve for moles of AgNO₃. Moles of AgNO₃ = (concentration of AgNO₃) * (volume of AgNO₃ solution) = (0.50 mol/L) * (1.0 L) = 0.50 mol AgNO₃

Since the stoichiometry of the balanced equation tells us that 2 moles of AgNO₃ are needed to react with 1 mole of BaCl₂, we can calculate the moles of AgNO₃ needed to react with 0.192 mol of BaCl₂: Moles of AgNO₃ needed = (2 moles AgNO₃ / 1 mole BaCl₂) * (0.192 mol BaCl₂) = 0.384 mol AgNO₃

Therefore, the moles of AgNO₃ in excess are the initial moles of AgNO₃ minus the moles of AgNO₃ needed: Moles of AgNO₃ in excess = 0.50 mol AgNO₃ - 0.384 mol AgNO₃ = 0.116 mol AgNO₃