What mass of AgN03 would you dissolve in water to get 1g of silver?
AgNO2 = 108 + 14 +3(16) = 170 g/mol
Ag = 108 g/mol
so AgNO3 is 108 grams silver/170 grams Agno3
1 gram silver(170 g AgNO3/108 g Ag)
= 1.57 grams
thanks!
To calculate the mass of AgNO3 needed to obtain 1g of silver, we need to understand the molar ratio between silver (Ag) and silver nitrate (AgNO3). The molar ratio is based on the balanced chemical equation of the reaction.
The balanced equation for the reaction between AgNO3 and Ag is:
2 AgNO3 → 2 Ag + O2 + 2 NO2
From the equation, we can see that 2 moles of AgNO3 will produce 2 moles of Ag.
To calculate the mass of AgNO3 needed, we need to know the molar mass of silver (Ag) and silver nitrate (AgNO3).
The molar mass of Ag is 107.87 g/mol.
The molar mass of AgNO3 is 169.87 g/mol.
Now, we can set up a dimensional analysis to find the mass of AgNO3:
(1 g Ag) × (1 mole AgNO3 / 2 moles Ag) × (169.87 g AgNO3 / 1 mole AgNO3) = x g AgNO3
Simplifying the equation, we get:
x = (1 g Ag) × (169.87 g AgNO3 / 2 × 107.87 g Ag)
x = 0.783 g AgNO3 (rounded to three decimal places)
Therefore, to obtain 1g of silver, you would need to dissolve approximately 0.783 grams of AgNO3 in water.