If 48.9g of barium chloride dihydrate, BaCl2.2H20, was treated with excess silver nitrate solution. (show formula)
a). how many moles of silver chloride precipitated?
b). how many grams of silver chloride precipitated?
thank you!
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
thank you for your response, I think I can solve the stoichiometry, the main issue for me is trying to set up the formula.
do you think could help me try to set up the formula? the components confuse me.
To determine the number of moles of silver chloride precipitated (AgCl), you need to first calculate the number of moles of barium chloride dihydrate (BaCl2.2H2O).
a) Let's start by calculating the molar mass of BaCl2.2H2O:
- Molar mass of Ba = 137.33 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of H2O = 18.02 g/mol
Since the compound contains two chloride ions and two water molecules, the molar mass of BaCl2.2H2O can be calculated as:
Molar mass of BaCl2.2H2O = (137.33 g/mol) + (2 × 35.45 g/mol) + (2 × 18.02 g/mol) = 244.26 g/mol
b) Now, let's calculate the number of moles of BaCl2.2H2O:
Number of moles = Mass / Molar mass
Number of moles of BaCl2.2H2O = 48.9 g / 244.26 g/mol = 0.2 mol
c) The chemical equation for the reaction is:
BaCl2.2H2O + 2AgNO3 -> 2AgCl + Ba(NO3)2 + 2H2O
From the balanced equation, we can see that 1 mole of BaCl2.2H2O will produce 2 moles of AgCl.
Therefore, the number of moles of AgCl precipitated will be:
Number of moles of AgCl = 2 × Number of moles of BaCl2.2H2O = 2 × 0.2 mol = 0.4 mol
To calculate the mass of AgCl precipitated, we need to know the molar mass of AgCl.
The molar mass of AgCl can be calculated as:
- Molar mass of Ag = 107.87 g/mol
- Molar mass of Cl = 35.45 g/mol
The molar mass of AgCl is:
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Finally, we can calculate the mass of AgCl precipitated:
Mass of AgCl = Number of moles of AgCl × Molar mass of AgCl
Mass of AgCl = 0.4 mol × 143.32 g/mol ≈ 57.33 g of AgCl
Therefore, the answer to part b) is approximately 57.33 grams of AgCl precipitated.