2. When exactly 61.9 g of silver nitrate react with magnesium, how many grams of Ag are prepared?
Balanced equation:
Step 1. Write a balanced equation.
Mg + 2AgNO3 ==> 2Ag + Mg(NO3)2
Step 2. Convert what you have (61.9 g AgNO3) to mols.
# mols = grams/molar mass
61.9/170 = 0.36
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (mols AgNO3) to mols of what you want (mols Mg).
mols Mg = mols AgNO3 x (1 mol Mg/2 mol AgNO3) = 0.36 x (1/2) = 0.18 mols Mg
Step 4. Now convert mols Mg to grams.
g Mg = mols Mg x molar mass Mg.
I have used estimated molar masses so you need to redo this using exact numbers. Print this for your records. This is a good general procedure for stoichiometric problems. You will use this many times. Check my arithmetic.
To determine the mass of silver (Ag) prepared when 61.9 g of silver nitrate (AgNO3) reacts with magnesium (Mg), you need to:
1. Write out the balanced equation for the reaction:
2 AgNO3 + Mg -> Mg(NO3)2 + 2 Ag
2. Determine the molar mass of silver nitrate (AgNO3):
AgNO3 = (1 Ag x 1) + (1 N x 14.01) + (3 O x 16.00)
= 107.87 g/mol
3. Convert the given mass of silver nitrate to moles:
Moles of AgNO3 = 61.9 g / 107.87 g/mol
4. Use the stoichiometry from the balanced equation to find the moles of silver (Ag) produced:
Moles of Ag = Moles of AgNO3 x (2 moles of Ag / 2 moles of AgNO3)
= Moles of AgNO3
5. Convert moles of silver (Ag) to grams:
Grams of Ag = Moles of Ag x Molar mass of Ag
By following these steps, you can determine the mass of silver (Ag) that is prepared when 61.9 g of silver nitrate reacts with magnesium.