The balanced chemical equation is:
MgCl2 (aq) + 2AgNO3 (aq) → Mg(NO3)2 (aq) + 2AgCl (s)
To find the mass of AgCl produced, we need to use stoichiometry. First, we need to find the limiting reactant, which is the one that runs out first. We can do this by converting the given masses and volumes into moles using molar mass and molarity.
Molar mass of MgCl2 = 24.3 + 2(35.5) = 95.3 g/mol Moles of MgCl2 = 15.0 g / 95.3 g/mol = 0.157 mol
Molarity of AgNO3 = 1.5 M Moles of AgNO3 = 1.5 M x 0.025 L = 0.0375 mol
Next, we need to compare the mole ratios of the reactants using the coefficients in the balanced equation.
Mole ratio of MgCl2 to AgNO3 = 1:2 Moles of AgNO3 needed for complete reaction with MgCl2 = 0.157 mol x 2 = 0.314 mol
Since we have less moles of AgNO3 than needed (0.0375 mol < 0.314 mol), AgNO3 is the limiting reactant.
Finally, we need to use the mole ratio of AgNO3 to AgCl to find the moles of AgCl produced.
Mole ratio of AgNO3 to AgCl = 1:1 Moles of AgCl produced = moles of AgNO3 used = 0.0375 mol
To find the mass of AgCl produced, we need to multiply by its molar mass.
Molar mass of AgCl = 107.9 + 35.5 = 143.4 g/mol Mass of AgCl produced = moles of AgCl x molar mass of AgCl = 0.0375 mol x 143.4 g/mol = 5.38 g
So, 5.38 g of silver chloride will be produced in this reaction.
Received message. This is a precipitation reaction where magnesium chloride and silver nitrate react to form solid silver chloride and magnesium nitrate solution . The balanced chemical equation is: MgCl2 (aq) + 2AgNO3 (aq) → Mg(NO3)2 (aq) + 2AgCl (s) To find the mass of AgCl produced, we need to use stoichiometry. First, we need to find the limiting reactant, which is the one that runs out first. We can do this by converting the given masses and volumes into moles using molar mass and molarity. Molar mass of MgCl2 = 24.3 + 2(35.5) = 95.3 g/mol Moles of MgCl2 = 15.0 g / 95.3 g/mol = 0.157 mol Molarity of AgNO3 = 1.5 M Moles of AgNO3 = 1.5 M x 0.025 L = 0.0375 mol Next, we need to compare the mole ratios of the reactants using the coefficients in the balanced equation. Mole ratio of MgCl2 to AgNO3 = 1:2 Moles of AgNO3 needed for complete reaction with MgCl2 = 0.157 mol x 2 = 0.314 mol Since we have less moles of AgNO3 than needed (0.0375 mol < 0.314 mol), AgNO3 is the limiting reactant. Finally, we need to use the mole ratio of AgNO3 to AgCl to find the moles of AgCl produced. Mole ratio of AgNO3 to AgCl = 1:1 Moles of AgCl produced = moles of AgNO3 used = 0.0375 mol To find the mass of AgCl produced, we need to multiply by its molar mass. Molar mass of AgCl = 107.9 + 35.5 = 143.4 g/mol Mass of AgCl produced = moles of AgCl x molar mass of AgCl = 0.0375 mol x 143.4 g/mol = **5.38 g** So, **5.38 g** of silver chloride will be produced in this reaction.