Question

When silver nitrate (169.87 g/mol) reacts with barium chloride (mm= 208.23 g/mol), silver chloride (mm=143.32 g/mol) and barium nitrate (mm=261.337 g/mol) are formed. how many grams of silver chloride are formed when 10.8 g of silver nitrate reacts with 15.0 g of barium chloride?

Answer 1

The mass of silver chloride formed when 10.8 g of silver nitrate reacts with 15.0 g of barium chloride is 9.12 g.

Step-by-step explanation:

To find the mass of silver chloride formed when 10.8 g of silver nitrate reacts with 15.0 g of barium chloride, we need to determine the limiting reactant and use stoichiometry to calculate the mass of silver chloride formed.

First, we need to find the number of moles of silver nitrate and barium chloride:

10.8 g AgNO₃ * (1 mol AgNO₃ / 169.87 g AgNO₃) = 0.0636 mol AgNO₃

15.0 g BaCl₂ * (1 mol BaCl₂ / 208.23 g BaCl₂) = 0.0720 mol BaCl₂

Next, we need to determine the stoichiometric ratio between silver nitrate and silver chloride:

AgNO₃ + BaCl₂ → AgCl + Ba(NO₃)₂

1 mol AgNO₃ produces 1 mol AgCl

So, the number of moles of AgCl formed will be the same as the moles of AgNO₃, which is 0.0636 mol.

Finally, we can calculate the mass of AgCl by multiplying the number of moles by the molar mass:

0.0636 mol AgCl * 143.32 g/mol AgCl = 9.12 g AgCl