Question

Copper reacts with silver nitrate through single replacement. If 2.75 g of silver are produced from the reacti?If 2.75 g of silver are produced from the reaction, how many moles of copper(II) nitrate are also produced? How many moles of each reactant are required in this reaction?

Answer 1

`\boxed{\text{0.0128 mol Cu(NO$_(3)$)$_(2)$; 0.0128 mol Cu; 0.0225 mol AgNO$_(3)$}}`

Step-by-step 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

`n = \text{2.75 g Ag} * \frac{\text{1 mol Ag}}{\text{107.8 g Ag}} = \text{0.025 51 mol Ag}`

(ii) Calculate the moles of Cu(NO₃)₂

The molar ratio is 1 mol Cu(NO₃)₂:2 mol Ag

`n = \text{0.025 51 mol Ag}* \frac{\text{1 mol Cu(NO$_(3)$)$_(2)$}}{\text{2 mol Ag}} = \boxed{\textbf{0.0128 mol Cu(NO$_(3)$)$_(2)$}}`

(c) Moles of Cu

The molar ratio is 1 mol Cu:2 mol Ag

`n = \text{0.025 51mol Ag} * \frac{\text{1 mol Cu}}{\text{2 mol Ag}}= \boxed{\textbf{0.0128 mol Cu}}`

(d) Moles of AgNO₃

The molar ratio is 2 mol Ag:2 mol AgNO₃

`n = \text{0.025 51 mol Ag} * \frac{\text{2 mol AgNO$_(3)$}}{\text{2 mol Ag}}= \boxed{\textbf{0.0255 mol AgNO$_(3)$}}`