Question

Consider the balanced chemical reaction below. What is the maximum number of grams of silver chloride that can be formed if a solution containing 18.0 g of silver nitrate and 32.4 g of iron(III) chloride are mixed?

Answer 1

Answer: 15.2 grams

Step-by-step explanation:

The balanced chemical equation is :

`FeCl_3(aq)+3AgNO_3(aq)\rightarrow 3AgCl(s)+Fe(NO_3)_3(aq)`

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} FeCl_3=(32.4g)/(162.2g/mol)=0.200moles`

`\text{Moles of} AgNO_3=(18.0g)/(169.87g/mol)=0.106moles`

According to stoichiometry :

3 moles of
`AgNO_3` require = 1 mole of
`FeCl_3`

Thus 0.106 moles of
`AgNO_3` will require=
`(1)/(3)* 0.106=0.0353moles` of
`FeCl_3`

Thus
`AgNO_3` is the limiting reagent as it limits the formation of product and
`FeCl_3` is the excess reagent.

As 3 moles of
`AgNO_3` give = 3 moles of
`AgCl`

Thus 0.106 moles of
`AgNO_3` give =
`(3)/(3)* 0.106=0.106moles` of
`AgCl`

Mass of
`AgCl=moles* {\text {Molar mass}}=0.106moles* 143.32g/mol=15.2g`

Thus 15.2 g of
`AgCl` will be produced from the given masses of both reactants.