Potassium chloride reacts with silver nitrate to produce solid silver chloride and potassium nitrate: KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq) How many grams of KCl are required to react completely with 250 mL of 0.030 M AgNO3?
The Correct Answer and Explanation is:
To find how many grams of KCl are required to react completely with 250 mL of 0.030 M AgNO3, we need to follow these steps:
Step 1: Write the balanced chemical equation
The reaction is:KCl(aq)+AgNO3(aq)→AgCl(s)+KNO3(aq)\text{KCl} (aq) + \text{AgNO}_3 (aq) \rightarrow \text{AgCl} (s) + \text{KNO}_3 (aq)KCl(aq)+AgNO3(aq)→AgCl(s)+KNO3(aq)
From this equation, you can see that 1 mole of KCl reacts with 1 mole of AgNO3.
Step 2: Calculate the moles of AgNO3
We are given the volume and molarity of AgNO3:
- Volume of AgNO3 = 250 mL = 0.250 L
- Molarity of AgNO3 = 0.030 M
The number of moles of AgNO3 is given by the formula:moles of AgNO3=Molarity×Volume (in L)\text{moles of AgNO}_3 = \text{Molarity} \times \text{Volume (in L)}moles of AgNO3=Molarity×Volume (in L)moles of AgNO3=0.030 mol/L×0.250 L=0.0075 mol\text{moles of AgNO}_3 = 0.030 \, \text{mol/L} \times 0.250 \, \text{L} = 0.0075 \, \text{mol}moles of AgNO3=0.030mol/L×0.250L=0.0075mol
Step 3: Moles of KCl required
From the balanced equation, we know that the reaction occurs in a 1:1 ratio, so the moles of KCl required will also be 0.0075 mol.
Step 4: Convert moles of KCl to grams
Now, we need to find the mass of KCl required. The molar mass of KCl is:Molar mass of KCl=39.1 g/mol (for K)+35.5 g/mol (for Cl)=74.6 g/mol\text{Molar mass of KCl} = 39.1 \, \text{g/mol (for K)} + 35.5 \, \text{g/mol (for Cl)} = 74.6 \, \text{g/mol}Molar mass of KCl=39.1g/mol (for K)+35.5g/mol (for Cl)=74.6g/mol
The mass of KCl required is:mass of KCl=moles of KCl×molar mass of KCl\text{mass of KCl} = \text{moles of KCl} \times \text{molar mass of KCl}mass of KCl=moles of KCl×molar mass of KClmass of KCl=0.0075 mol×74.6 g/mol=0.5595 g\text{mass of KCl} = 0.0075 \, \text{mol} \times 74.6 \, \text{g/mol} = 0.5595 \, \text{g}mass of KCl=0.0075mol×74.6g/mol=0.5595g
Final Answer:
So, 0.56 grams of KCl are required to react completely with 250 mL of 0.030 M AgNO3.
