When solutions of AgNO3 and NaOH react, the balanced molecular equation is as follows:
2AgNO3(aq) + 2NaOH(aq) → Ag2O(s) + 2NaNO3(aq) + H2O(l)
How much Ag2O is produced when 2.00 g of AgNO3 and 0.300 g of NaOH react?
To determine the amount of Ag2O produced, we need to convert the given masses of AgNO3 and NaOH to moles, and then use the stoichiometry of the balanced equation.
Step 1: Convert the mass of AgNO3 to moles.
molar mass of AgNO3 = atomic mass of Ag + atomic mass of N + (3 x atomic mass of O)
= (107.87 g/mol) + (14.01 g/mol) + (3 x 16.00 g/mol)
= 169.87 g/mol
moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 2.00 g / 169.87 g/mol
= 0.0118 mol
Step 2: Convert the mass of NaOH to moles.
molar mass of NaOH = atomic mass of Na + atomic mass of O + atomic mass of H
= (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol)
= 40.00 g/mol
moles of NaOH = mass of NaOH / molar mass of NaOH
= 0.300 g / 40.00 g/mol
= 0.0075 mol
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the moles of AgNO3 and NaOH in a 1:1 ratio. Since the ratio is the same, the limiting reactant is the one with fewer moles. In this case, that is NaOH with 0.0075 mol.
Step 4: Determine the amount of Ag2O produced using stoichiometry.
From the balanced equation, we see that 2 moles of AgNO3 will produce 1 mole of Ag2O.
moles of Ag2O = (0.0075 mol NaOH) x (1 mol Ag2O / 2 mol NaOH)
= 0.00375 mol
Step 5: Convert moles of Ag2O to grams.
molar mass of Ag2O = (2 x atomic mass of Ag) + atomic mass of O
= (2 x 107.87 g/mol) + 16.00 g/mol
= 231.74 g/mol
grams of Ag2O = moles of Ag2O x molar mass of Ag2O
= 0.00375 mol x 231.74 g/mol
= 0.869 g
Therefore, 0.869 grams of Ag2O is produced when 2.00 grams of AgNO3 and 0.300 grams of NaOH react.