2AgNO3 (s) + MgCl2 (aq) - -> 2AgCl (s) + Mg (NO3) 2 (aq) The student starts with 50.0 grams of mgcl2 and an excess of AgNO3. If she makes 125.1g of AgCl, what is her percent yield?

The percent yield is 83.12%

calculation

percent yield = actual yield / theoretical yield x 100

actual yield = 125.1 g

Theoretical yield is calculated as below

Step 1: find the moles of MgCl₂

moles = mass: molar mass

from periodic table the molar mass of mgCl₂ = 24.3 + (35.5 x2) = 95.3 g/mol

moles = 50.0 g: 95.3 g/mol = 0.525 moles

Step 2: use the mole ratio to determine the moles of AgCl

from given equation MgCl₂: AgCl is 1:2

therefore the moles of AgCl = 0.525 x2/1 = 1.05 moles

Step 3: find the theoretical mass AgCl

mass = moles x molar mass

from periodic table the molar mass of Agcl = 107.87 + 35.5 = 143.37 g/mol

mass = 1.05 moles x 143.37 g/mol = 150.5 grams

The percent yield is therefore = (125.1 g/150.5) x 100 = 83.12%