AgNO3 (aq) + HCl (aq) �¨ AgCl (s) + HNO3 (aq)
40.30 grams of silver nitrate and excess HCl, how much silver chloride could you make?
Here is an example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine how much silver chloride (AgCl) could be produced from 40.30 grams of silver nitrate (AgNO3), we need to use the concept of stoichiometry and perform some calculations.
In the balanced chemical equation you provided:
AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq)
The stoichiometric ratio between AgNO3 and AgCl is 1:1. This means that for every 1 mole of AgNO3, 1 mole of AgCl is produced.
To find the amount of AgCl produced, we need to convert the given mass of AgNO3 to moles, and then use the stoichiometric ratio to determine the moles of AgCl.
First, we need to calculate the molar mass of AgNO3:
Ag = 107.87 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (x3)
Total molar mass of AgNO3 = (107.87 + 14.01 + (16.00 x 3)) g/mol = 169.87 g/mol
Now, we can calculate the moles of AgNO3:
moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
moles of AgNO3 = 40.30 g / 169.87 g/mol ≈ 0.2375 mol
Since the stoichiometric ratio is 1:1 between AgNO3 and AgCl, we have 0.2375 moles of AgCl produced as well.
Finally, we can calculate the mass of AgCl:
mass of AgCl = moles of AgCl × molar mass of AgCl
mass of AgCl = 0.2375 mol × (107.87 + 35.45) g/mol (Ag + Cl)
mass of AgCl ≈ 29.59 g
Therefore, approximately 29.59 grams of silver chloride could be produced from 40.30 grams of silver nitrate.