Answer:
[tex]\boxed{\text{0.0128 mol Cu(NO$_{3}$)$_{2}$; 0.0128 mol Cu; 0.0225 mol AgNO$_{3}$}}[/tex]
Explanation:
a) Balanced equation
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 63.55 169.87 187.56 107.87
Cu + 2AgNO₃ ⟶ Cu(NO₃)₂ + 2Ag
m/g: 2.75
(b) Moles of Cu(NO₃)₂
(i) Calculate the moles of Ag
[tex]n = \text{2.75 g Ag} \times \dfrac{\text{1 mol Ag}}{\text{107.8 g Ag}} = \text{0.025 51 mol Ag}[/tex]
(ii) Calculate the moles of Cu(NO₃)₂
The molar ratio is 1 mol Cu(NO₃)₂:2 mol Ag
[tex]n = \text{0.025 51 mol Ag}\times \dfrac{\text{1 mol Cu(NO$_{3}$)$_{2}$}}{\text{2 mol Ag}} = \boxed{\textbf{0.0128 mol Cu(NO$_{3}$)$_{2}$}}[/tex]
(c) Moles of Cu
The molar ratio is 1 mol Cu:2 mol Ag
[tex]n = \text{0.025 51mol Ag} \times \dfrac{\text{1 mol Cu}}{\text{2 mol Ag}}= \boxed{\textbf{0.0128 mol Cu}}[/tex]
(d) Moles of AgNO₃
The molar ratio is 2 mol Ag:2 mol AgNO₃
[tex]n = \text{0.025 51 mol Ag} \times \dfrac{\text{2 mol AgNO$_{3}$}}{\text{2 mol Ag}}= \boxed{\textbf{0.0255 mol AgNO$_{3}$}}[/tex]
