Question

Consider the following unbalanced equation: AgNO3 (aq) CaCl2 (aq) → AgCl (s) Ca(NO3)2 (aq) How many grams of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride?

Answer 1

Answer: 3 grams

Explanation:-

To calculate the number of moles for given molarity, we use the equation:

`\text{Moles of solute}={\text{Molarity of the solution}}*{\text{Volume of solution (in L)}}` .....(1)

Molarity of
`AgNO_3` solution = 0.20 M

Volume of
`AgNO_3` solution = 100.0 mL = 0.1 L

Putting values in equation 1, we get:

`\text{Moles of} AgNO_3={0.20}*{0.1}=0.02moles`

Molarity of
`CaCl_2` solution = 0.15 M

Volume of
`CaCl_2` solution = 100.0 mL = 0.1 L

Putting values in equation 1, we get:

`\text{Moles of} CaCl_2={0.15}*{0.1}=0.015moles`

The balanced chemical equation for the reaction is:

`2AgNO_3(aq)+CaCl_2(aq)\rightarrow 2AgCl(s)+Ca(NO_3)_2(aq)`

According to stoichiometry :

2 moles of
`AgNO_3` require 1 mole of
`CaCl_2`

Thus 0.02 moles of
`AgNO_3` will require=
`(1)/(2)* 0.02=0.01moles` of
`CaCl_2`

Thus
`AgNO_3` is the limiting reagent as it limits the formation of product and
`CaCl_2` is the excess reagent.

As 2 mole of
`AgNO_3` give = 2 mole of
`AgCl`

Thus 0.02 moles of
`AgNO_3` give =
`(1)/(1)* 0.02=0.02moles` of
`AgCl`

Mass of
`AgCl=moles* {\text {Molar mass}}=0.02moles* 143.32g/mol=3g`

Thus 3 g of
`AgCl` will be produced from the given masses of both reactants.