caculate the mass of agcl precipitated when 0.1g of nacl and 0.1g of agno3 are mixed in solution
Convert 0.1 g NaCl and 0.1 g AgNO3 (note: nacl and agno3 don't mean a thing--don't be so lazy in chemistry that you can't punch the CAPS key). Here is a worked example of a stoichiometry problem. Print this out--it will solve 99.9% of the stoichiometry problems you are likely to encounter.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the mass of AgCl precipitated when NaCl and AgNO3 are mixed in solution, we need to determine the limiting reactant first. The limiting reactant is the reactant that gets completely consumed, limiting the amount of product formed.
Step 1: Write the balanced chemical equation for the reaction:
AgNO3 + NaCl → AgCl + NaNO3
Step 2: Calculate the number of moles for each reactant:
Moles of NaCl = mass of NaCl / molar mass of NaCl
Moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
The molar masses for NaCl and AgNO3 are:
NaCl: 22.99 g/mol (sodium has atomic mass 22.99 and chlorine has atomic mass 35.45)
AgNO3: 169.87 g/mol (silver has atomic mass 107.87, nitrogen has atomic mass 14.01, and three oxygen atoms have atomic mass 16.00)
For our given masses:
Moles of NaCl = 0.1 g / 22.99 g/mol ≈ 0.00435 mol
Moles of AgNO3 = 0.1 g / 169.87 g/mol ≈ 0.00059 mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, compare the moles of each reactant using the stoichiometric ratio from the balanced equation.
Based on the balanced equation, the ratio between AgNO3 and NaCl is 1:1.
Since we have more moles of NaCl (0.00435 mol) compared to AgNO3 (0.00059 mol), AgNO3 is the limiting reactant.
Step 4: Determine the moles of AgCl formed:
Since AgNO3 is the limiting reactant, the moles of AgCl formed will be equal to the moles of AgNO3 used.
Moles of AgCl = Moles of AgNO3 = 0.00059 mol
Step 5: Calculate the mass of AgCl formed:
Mass of AgCl = Moles of AgCl × Molar mass of AgCl
The molar mass of AgCl is 143.32 g/mol (silver has atomic mass 107.87, chlorine has atomic mass 35.45).
Mass of AgCl = 0.00059 mol × 143.32 g/mol ≈ 0.0847 g
Therefore, approximately 0.0847 grams of AgCl will precipitate when 0.1 grams of NaCl and 0.1 grams of AgNO3 are mixed in a solution.