To find the molar mass of silver chloride (AgCl), we first need to determine how much AgCl can be produced from the given solutions of silver nitrate (AgNO3) and calcium chloride (CaCl2). The relevant reaction between these two compounds is:
\[ \text{AgNO}_3 (aq) + \text{CaCl}_2 (aq) \rightarrow \text{AgCl} (s) + \text{Ca(NO}_3)_2 (aq) \]
Step 1: Calculate the moles of reactants.
Moles of AgNO3:
- Volume of AgNO3 solution = 200.0 mL = 0.200 L
- Molarity of AgNO3 = 0.30 M
\[ \text{Moles of AgNO}_3 = \text{Molarity} \times \text{Volume(L)} = 0.30 , \text{mol/L} \times 0.200 , \text{L} = 0.060 , \text{mol} \]
Moles of CaCl2:
- Volume of CaCl2 solution = 180.0 mL = 0.180 L
- Molarity of CaCl2 = 0.28 M
\[ \text{Moles of CaCl}_2 = \text{Molarity} \times \text{Volume(L)} = 0.28 , \text{mol/L} \times 0.180 , \text{L} = 0.0504 , \text{mol} \]
Step 2: Determine the limiting reactant.
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of CaCl2 to produce 1 mole of AgCl.
- Moles of AgNO3 available = 0.060 mol
- Moles of CaCl2 available = 0.0504 mol
Since both reactants react in a 1:1 ratio, the limiting reactant is CaCl2 because there are fewer moles available (0.0504 mol).
Step 3: Calculate the moles of AgCl produced.
Since CaCl2 is the limiting reactant, the moles of AgCl produced will be equal to the moles of CaCl2:
\[ \text{Moles of AgCl produced} = 0.0504 , \text{mol} \]
Step 4: Calculate the molar mass of AgCl.
To find the molar mass of AgCl, we can sum the molar masses of silver (Ag) and chlorine (Cl):
- Molar mass of Ag = 107.87 g/mol
- Molar mass of Cl = 35.45 g/mol
\[ \text{Molar mass of AgCl} = 107.87 , \text{g/mol} + 35.45 , \text{g/mol} = 143.32 , \text{g/mol} \]
Conclusion
The molar mass of silver chloride (AgCl) is 143.32 g/mol.