Silver nitrate and aluminum chloride react with each other by exchanging anions:

Question

Silver nitrate and aluminum chloride react with each other by exchanging anions:

3AgNO3 AlCl3 --> Al(NO3)3 3 AgCl

What mass in grams of AlCl is produced when 4.22g. of AgNO3 react with 7.73g. of AlCl3?

Thank you!!

Answer 1

3AgNO3 &  AlCl3 --> Al(NO3)3 &   3 AgCl

3 moles AgNO3 @ 169.88 g/mol = 509.64 grams AgNO3 react

1 mole of AlCl3 @ 133.34 g/mol = 133.34 grams of AlCl3 react

the ratio that reacts is 509.64 g AgNO3 / 133.34 g AlCL3 = 3.82

the ratio that they used was 4.22g AgNO3 / 7.73 gAlCl3 = 0.546

they have seriously limited the amount of AgNO3, it is  your "limiting reagent"

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"What mass in grams of AlCl is produced when 4.22g. of AgNO3 react"

now I think that you meant:

What mass in grams of AgCl is produced when 4.22g. of AgNO3 react

if so, then the balanced equation produces 3 moles of AgCl @ 143.32 g/mol = 429.96 grams of AgCl are produced from 509.64 g AgNO3 referred to earlier

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so:

4,22 g AgNO3 @ 429.96 g AgCl / 509.64 g AgNO3 = 3.56 grams of AgCl

your answer is : 3.56g of AgCL can be produced from 4.22 grams of AgNO3

Answer 2

from the chemical equation 3 moles of AgNO3 react with 1 mole of AlCl3 to produce 3 moles of AgCl

so moles AgNO3 = moles AgCl

moles AgNO3 = weight/ molecular weight

= 4.22/169.9 =0.025 mole

so moles AgCl=0.025 mole

moles= weight/ molecular weight>>> weight= moles X molecular weight

weight of  AgCl =  0.025  X  143.4

=3.585 gm

hope its clearful to you...