What mass of silver chloride, AgCl will precipitate from a solution that contains 1.50 g of, calcium chloride, CaCl if an excess amount of silver nitrate is present?
CaCl2 + 2AgNO3 --> 2AgCl+ Ca(NO3)2
I keep getting 5!but its wrong...right?
CaCl2 is the limiting reagent. 1 mole of CaCl2 =2 moles of AgNO3=2 moles of AgCl =1 mole of Ca(NO3)2.
1.50g of CaCl2/110.98 g/mol of CaCl2= number of moles of CaCl2
number of moles of CaCl2*(2 moles of AgCl/1 mole of CaCl2) * 143.32 g of AgCl*mole-1= mass of silver chloride.
Didn't punch in the number, but this should be right.
you idiot you didnt write the number
To find the mass of silver chloride (AgCl) that will precipitate from the given solution, we need to use stoichiometry.
First, let's calculate the number of moles of calcium chloride (CaCl2) in the solution.
Given:
Mass of CaCl2 = 1.50 g
Molar mass of CaCl2 = 40.08 g/mol
Number of moles of CaCl2 = mass/molar mass
= 1.50 g/40.08 g/mol
= 0.0374 mol
According to the balanced chemical equation:
1 mole of CaCl2 reacts with 2 moles of AgCl.
So, the number of moles of AgCl that will precipitate is twice the number of moles of CaCl2.
Number of moles of AgCl = 2 x 0.0374 mol
= 0.0748 mol
Finally, let's calculate the mass of AgCl precipitated.
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = number of moles x molar mass
= 0.0748 mol x 107.87 g/mol
= 8.07 g
Therefore, the mass of silver chloride (AgCl) that will precipitate from the solution is 8.07 grams.
To determine the mass of silver chloride precipitated from a solution containing calcium chloride, we need to calculate the stoichiometry of the reaction.
The balanced chemical equation for the reaction between calcium chloride (CaCl2) and silver nitrate (AgNO3) is:
CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2
From the equation, we can see that for every 1 mole of calcium chloride, 2 moles of silver chloride are produced.
Step 1: Calculate the number of moles of calcium chloride (CaCl2):
Given: Mass of calcium chloride = 1.50 g
Molar mass of CaCl2 = 40.08 g/mol + (2 × 35.45 g/mol) = 110.98 g/mol
Number of moles = Mass / Molar mass = 1.50 g / 110.98 g/mol ≈ 0.0135 mol
Step 2: Calculate the mass of silver chloride (AgCl) produced:
From the stoichiometry of the reaction, we know that 2 moles of AgCl are produced for every 1 mole of CaCl2.
So, the molar ratio of AgCl to CaCl2 is 2:1.
Mass of AgCl = Number of moles of CaCl2 × Molar mass of AgCl
Number of moles of AgCl = 2 × Number of moles of CaCl2 = 2 × 0.0135 mol = 0.0270 mol
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = 0.0270 mol × 107.87 g/mol ≈ 2.92 g
Therefore, approximately 2.92 grams of silver chloride (AgCl) will precipitate from the solution containing 1.50 grams of calcium chloride (CaCl2).