In a chemical analysis, 3.40g of silver nitrate in solution reacted with excess sodium chloride to produce 2.81g of precipitate. What is the percent yield? Please show work. Thanks
To calculate the percent yield, you need to compare the amount of the precipitate obtained to the theoretical yield, which is the amount of precipitate that should have been produced according to the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl):
AgNO3 + NaCl -> AgCl + NaNO3
From the equation, we can see that the ratio of AgNO3 to AgCl is 1:1. This means that 1 mole of AgNO3 should produce 1 mole of AgCl.
Now, we need to calculate the theoretical yield of AgCl. To do this, we need to convert the mass of AgNO3 used into moles, and then use the mole ratio to determine the moles of AgCl produced.
First, let's calculate the moles of AgNO3:
molar mass of AgNO3 = atomic mass of Ag + atomic mass of N + 3 * atomic mass of O
= (107.87 g/mol) + (14.01 g/mol) + 3 * (16.00 g/mol)
≈ 169.87 g/mol.
moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 3.40 g / 169.87 g/mol
≈ 0.02 mol.
Since the mole ratio of AgNO3 to AgCl is 1:1, the moles of AgCl formed will also be 0.02 mol.
Now we can calculate the mass of AgCl formed using the molar mass of AgCl:
molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= (107.87 g/mol) + (35.45 g/mol)
≈ 143.32 g/mol.
mass of AgCl = moles of AgCl * molar mass of AgCl
= 0.02 mol * 143.32 g/mol
= 2.87 g.
Now we can calculate the percent yield:
percent yield = (actual yield (g) / theoretical yield (g)) * 100%
= (2.81 g / 2.87 g) * 100%
≈ 97.9%.
Therefore, the percent yield is approximately 97.9%.