find the mass of silver chloride formed from 33.2ml of a 0.100M solution of silver nitrate and 200ml of 0.200M solution of calcium chloride solution
Solution
plan:use mass to mass calculations
mass of AgNO3-moles of AgNO3-moles of AgCl-mass of AgCl
0.1mol/1L*0.0332L+0.200mol/1L=0.04332mol(moles of calcium and moles of silver nitrate)
0.04332moles*2molesAgCl/(1moleCaCl2*2molAg(NO3)2)=0.04332moles Agcl
0.04332molsAgCl*143.45g/1mole
=6.211g of AgCl
To find the mass of silver chloride formed, we can use mass-to-mass calculations. Here's how you can calculate it step by step:
1. Calculate the moles of silver nitrate (AgNO3) in the 33.2 mL of the 0.100 M solution:
Moles of AgNO3 = (0.100 mol/L) * (0.0332 L) = 0.00332 mol
2. Calculate the moles of calcium chloride (CaCl2) in the 200 mL of the 0.200 M solution:
Moles of CaCl2 = (0.200 mol/L) * (0.200 L) = 0.0400 mol
3. Since silver nitrate, AgNO3, reacts with calcium chloride, CaCl2, in a 2:1 ratio to form silver chloride (AgCl), we need to determine the moles of AgCl formed.
Moles of AgCl = Moles of CaCl2 * (2 mol AgCl / 1 mol CaCl2) = 0.0400 mol * (2 mol AgCl / 1 mol CaCl2) = 0.0800 mol
4. Finally, convert the moles of AgCl to grams using the molar mass of AgCl:
Mass of AgCl = Moles of AgCl * Molar Mass of AgCl = 0.0800 mol * 143.45 g/mol = 11.476 g
Therefore, the mass of silver chloride formed from the given solutions is approximately 11.476 grams.
To find the mass of silver chloride formed, we will use mass-to-mass calculations.
Step 1: Calculate the moles of silver nitrate (AgNO3) and calcium chloride (CaCl2) used.
The volume of the silver nitrate solution is 33.2 mL, which is equivalent to 0.0332 L. The concentration of the silver nitrate solution is 0.100 M.
1 L of 0.100 M solution contains 0.100 mol of AgNO3.
Therefore, 0.0332 L of 0.100 M solution contains 0.100 mol/1 L * 0.0332 L = 0.00332 mol of AgNO3.
The volume of the calcium chloride solution is 200 mL, which is equivalent to 0.200 L. The concentration of the calcium chloride solution is 0.200 M.
1 L of 0.200 M solution contains 0.200 mol of CaCl2.
Therefore, 0.200 L of 0.200 M solution contains 0.200 mol/1 L * 0.200 L = 0.040 mol of CaCl2.
Step 2: Calculate the moles of silver chloride (AgCl) formed.
From the balanced chemical equation, we know that 2 moles of AgCl are formed for every 1 mole of CaCl2 and 2 moles of AgNO3 reacted.
Therefore, the moles of AgCl formed would be:
0.04332 mol * (2 mol AgCl / (1 mol CaCl2 * 2 mol Ag(NO3)2)) = 0.04332 mol.
Step 3: Calculate the mass of AgCl formed.
The molar mass of AgCl is 143.45 g/mol.
Therefore, the mass of AgCl formed would be:
0.04332 mol * 143.45 g/1 mol = 6.211 g.
So, the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and 200 mL of a 0.200 M solution of calcium chloride is 6.211 grams.