# How many atoms of each element of AgNO_3 are in 0.15 mol of AgNO_3?

Jun 29, 2018

$9.0 \text{x"10^22"Ag atoms}$

$9.0 \text{x"10^22"N atoms}$

$2.7 \text{x"10^23"O atoms}$

#### Explanation:

1. From 0.15mol of $A g N {O}_{3}$, we can find the moles of each element by counting how many of each element is present in the entire compound. In $A g N {O}_{3}$, there's 1 Ag, 1 N and 3 O.
2. Then, we convert from moles of each element to the number of atoms using Avogadro's number $6.02 \text{x} {10}^{23}$.

I went ahead and set up the two steps above in 1 line for each element:
$\text{Ag atoms"=0.15 "mol " AgNO_3*(1 "mol " Ag)/(1 "mol " AgNO_3)*(6.02"x"10^23 "Ag atoms")/(1 "mol " Ag)=9.0"x"10^22"Ag atoms}$

$\text{N atoms"=0.15 "mol " AgNO_3*(1 "mol " N)/(1 "mol " AgNO_3)*(6.02"x"10^23 "N atoms")/(1 "mol " N)=9.0"x"10^22"N atoms}$

$\text{O atoms"=0.15 "mol " AgNO_3*(3 "mol " O)/(1 "mol " AgNO_3)*(6.02"x"10^23 "O atoms")/(1 "mol " O)=2.7"x"10^23"O atoms}$