if 30ml of .150 M CaCl2 is added to 22.5ml of .1 M AgNO3, what is the mass of the AgCl precipitate?
This is a limiting reagent problem. You know that because BOTH reactants are given.
Write the equation and balance it.
CaCl2 + 2AgNO3 ==> 2AgCl + Ca(NO3)2
Convert CaCl2 given to moles. moles = M x L.
Convert AgNO3 given to moles. Again, moles = M x L.
Using the coefficients in the balanced equation, convert moles CaCl2 to moles AgCl.
Same process, convert moles AgNO3 to moles CaCl2.
In limiting reagent problems usually these two values are not the same which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
Now, using the smaller value from above, convert mols AgCl to grams. g = moles x molar mass.
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To find the mass of the AgCl precipitate formed when 30 mL of 0.150 M CaCl2 is added to 22.5 mL of 0.1 M AgNO3, you'll need to use a stoichiometry calculation.
Step 1: Determine the equation for the reaction between CaCl2 and AgNO3.
CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl
Step 2: Calculate the number of moles of CaCl2 and AgNO3 used.
moles of CaCl2 = volume of CaCl2 (L) × concentration of CaCl2 (M)
moles of CaCl2 = 0.030 L × 0.150 M
moles of AgNO3 = volume of AgNO3 (L) × concentration of AgNO3 (M)
moles of AgNO3 = 0.0225 L × 0.100 M
Step 3: Determine the limiting reactant.
To find the limiting reactant, compare the ratios of the reactants in the balanced chemical equation to their moles obtained in Step 2. The reactant that produces the smaller amount of product is the limiting reactant.
moles of AgCl produced from CaCl2 = moles of CaCl2 × (2 moles of AgCl / 1 mole of CaCl2)
moles of AgCl produced from AgNO3 = moles of AgNO3 × (2 moles of AgCl / 2 moles of AgNO3)
Compare the moles of AgCl produced from both reactants. The reactant that produces less AgCl is the limiting reactant.
Step 4: Calculate the moles of AgCl produced.
moles of AgCl = moles of limiting reactant × (2 moles of AgCl / 1 mole of CaCl2 or 2 moles of AgNO3)
Step 5: Convert moles of AgCl to grams.
mass of AgCl = moles of AgCl × molar mass of AgCl
The molar mass of AgCl is equal to the atomic mass of silver (Ag) plus the atomic mass of chlorine (Cl). You can find these atomic masses on the periodic table.
Step 6: Calculate the mass of the AgCl precipitate.
Plug in the values obtained from the previous steps into the equation:
mass of AgCl = moles of AgCl × molar mass of AgCl