What mass of AgN03 would you dissolve in water to get 1g of silver?
1g Ag x (molar mass AgNO3/atomic mass Ag) = ?
What didn't you like about Damon's answer? I didn't see anything wrong with it; he may have rounded the numbers more than you wanted.
To determine the mass of AgNO3 required to obtain 1g of silver, we need to consider the molar mass and the stoichiometry of the reaction. The molar mass of silver (Ag) is approximately 107.87 g/mol, and the molar mass of AgNO3 is approximately 169.87 g/mol.
The balanced chemical equation for the reaction between AgNO3 and Ag is:
2AgNO3 → 2Ag + 2NO2 + O2
From the equation, we can see that 2 moles of AgNO3 produce 2 moles of Ag. This means that the ratio of AgNO3 to Ag is 1:1.
To calculate the mass of AgNO3 required, we can use the formula:
Mass of AgNO3 = (Mass of Ag) x (Molar mass of AgNO3) / (Molar mass of Ag)
Substituting the values, we have:
Mass of AgNO3 = (1g) x (169.87 g/mol) / (107.87 g/mol)
Simplifying the calculation, we get:
Mass of AgNO3 ≈ 1.57g
Therefore, approximately 1.57g of AgNO3 would need to be dissolved in water to obtain 1g of silver.