Question

40 ml of 0.1M cacl2 solution is mixed with 50 ml of 0.2 M Agno3 solution mole of agcl formed in rxn are

Answer 1

Answer:

80 mmol or 0.08 mol of AgCl

Explanation:

Let's first understand what is mmol or millimol. millimol is thousandth part of the mole.

When we multiply the concentration (molarity- number of moles of solute in 1 liter of solution) with litre we get number of moles. When we multiply molarity with milliliters we get millimole.

Let calculate the mmol of CaCl2 in 40 mL of 0.1 M solution

it would be 4 mmol

If we want to know how many Cl- is given by the CaCl solution we multiply the mmol by 2 because 1 CaCl2 gives 2 Cl- ion

So we have 8 mmol of Cl- ions

Similarly mmol of AgNO3 are 50*0.2= 10 mmol

The stoichiometric equation for the reaction is

$CaCl_2 + 2AgNO_3 -> 2AgCl +Ca(NO_3)_2$

We can see 1 mol of CaCl2 reacts with 2 mol of AgNO3

so if we have 4 mol of CaCl2 we will 8 mmol of AgCl

Whereas we have 10 mmol of AgCl

so CaCl2 is the limiting reagent.

1 mol of CaCl2 gives 2 mol AgCl

so 4 mmol will give 8 mmol of AgCl

Limiting reagent: The limiting reactant or limiting reagent is a reactant in a chemical reaction that determines the amount of product that is formed. Identification of the limiting reactant makes it possible to calculate the theoretical yield of a reaction.