Question

If an experiment with 10.2 g barium chloride produced 14.5 g silver chloride, calculate thr experimental mole ratio of silver chloride to barium chloride

Answer 1

Answer:

2.06/1

Explanation:

First, we will find the moles of each compound.

The molar mass of BaCl₂ is 208.23 g/mol. The moles corresponding to 10.2 g are:

10.2 g × (1 mol/208.23 g) = 0.0490 mol

The molar mass of AgCl is 143.32 g/mol. The moles corresponding to 14.5 g are:

14.5 g × (1 mol/143.32 g) = 0.101 mol

The experimental mole ratio of AgCl to BaCl₂ is 0.101 mol/0.0490mol = 2.06/1

This is very close to the theoretical mole ratio of AgCl to BaCl₂, according to the following reaction, which is 2/1.

BaCl₂ + 2 Ag⁺ → 2 AgCl + Ba²⁺

Answer 2

The  experimental  mole ratio  of silver chloride  to  barium chloride  is calculated as below

fin the mole of each compound

mole= mass/molar  mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102  moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol =  0.049  moles of BaCl2

find the  mole ratio  by dividing each mole with the smallest  mole(0.049)

AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore  the mole   ratio   AgCl to  BaCl2  is  2 :1