A 50.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCl(aq). The AgCl precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? I'm currently stuck and need help asap!!
To find the amount of precipitate formed, we need to use stoichiometry. First, let's write the balanced chemical equation for the reaction:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
From the balanced equation, we can determine the mole ratio between AgNO3 and AgCl:
1 mole AgNO3 : 1 mole AgCl
Given:
Volume of AgNO3 solution = 50.0 mL = 50.0/1000 L = 0.0500 L
Molarity of AgNO3 solution = 0.200 M
We can use the formula:
moles = concentration × volume
moles of AgNO3 = 0.200 mol/L × 0.0500 L = 0.0100 mol
Since the mole ratio between AgNO3 and AgCl is 1:1, the moles of AgCl formed will be the same as the moles of AgNO3.
moles of AgCl = 0.0100 mol
Now, we need to find the molar mass of AgCl to convert the moles into grams.
The molar mass of AgCl is:
(1 atom of Ag × atomic mass of Ag) + (1 atom of Cl × atomic mass of Cl)
(1 × 107.87 g/mol) + (1 × 35.45 g/mol) = 143.32 g/mol
Mass of AgCl = moles of AgCl × molar mass of AgCl
Mass of AgCl = 0.0100 mol × 143.32 g/mol
Mass of AgCl = 1.4332 g
Therefore, the expected mass of AgCl precipitate is 1.4332 grams.
To determine the number of grams of precipitate obtained, you need to use stoichiometry to calculate the amount of AgCl precipitate formed.
The balanced chemical equation for the reaction between AgNO3 and NaCl is:
AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)
From the equation, you can see that for every 1 mole of AgNO3, 1 mole of AgCl precipitate is formed. So, to calculate the moles of AgCl precipitate, you need to first calculate the moles of AgNO3 used.
Moles of AgNO3 = Molarity of AgNO3 x Volume of AgNO3 (in liters)
= 0.200 M x 0.0500 L
= 0.0100 moles of AgNO3
Since the reaction is 1:1 between AgNO3 and AgCl, the moles of AgCl precipitate formed will also be 0.0100 moles.
To calculate the mass of AgCl precipitate, you need to use its molar mass. The molar mass of AgCl is calculated as follows:
Ag (107.87 g/mol) + Cl (35.45 g/mol) = 143.32 g/mol
Mass of AgCl = Moles of AgCl x Molar mass of AgCl
= 0.0100 moles x 143.32 g/mol
= 1.4332 grams of AgCl
Therefore, you should obtain approximately 1.4332 grams of AgCl precipitate.
AgNO3 + NaCl ==> AgCl + NaNO3
Here is a sample stoichiometry problem i posted. Just follow the instructions. moles AgNO3 = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html