Question

2. When exactly 61.9 g of silver nitrate react with magnesium, how many grams of Ag are prepared?
Balanced equation:

Answer 1

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.