If 1144 grams of silver chloride were allowed to decompose, how many liters of chlorine gas would be produced? (assume STP conditions)

Silver chloride (AgCl) decomposes according to the following reaction:

    `~2AgCl -gt 2Ag + ~Cl_2` ` `

Identify Information in the Problem:

This is a stoichiometry problem. The given amount is 1144 g AgCl. The final substance is L of `~Cl_2`.

Preliminary Calculation: Calculate the molar mass of AgCl.

Notice that the unit of the given substance (AgCl) is grams. Whenever the unit "grams" is used with a substance in a stoichometry problem, you will need to calculate the molar mass of the substance before you start working the problem. The molar mass of a compound is equal to the sum of the atomic masses of each element in the compound. The atomic mass of each element can be found in the periodic table.

   Atomic mass Ag = 107.868 g

   Atomic mass Cl = 35.453 g

   molar mass AgCl = 107.868 g + 35.453 g = 143.321 g

Stoichiometry Calculations:

Step 1: Convert the given substance to moles by multiplying by the conversion factor "1 mol = molar mass". Earlier, we determined that the molar mass of AgCl = 143.321 g. So, the conversion factor becomes, 1 mol AgCl = 143.321 g AgCl.

   1144 g AgCl x (1 mol AgCl/143.321 g AgCl) = 7.982 mol AgCl

Notice that the conversion factor "1 mol = 143.321 g" is oriented such that the "gram" part of the conversion factor is in the denominator. This enables us to cancel out grams and be left in moles. 

Step 2: Refer back to the balanced equation for the decomposition of AgCl. Notice that the coefficient of the given substance AgCl is 2, and the coefficient for the final substance `~Cl_2` is 1. Create a conversion factor using the coefficients of the given and final substances: 2 mol AgCl = 1 mol `~Cl_2` .

Multiply the moles of AgCl (from Step 1) by the coefficient conversion factor.

   7.982 mol AgCl x (1 mol `~Cl_2` /2 mol AgCl) = 3.991 mol `~Cl_2`

Notice that the coefficient conversion factor is oriented such that the "mol AgCl" part of the conversion factor is in the denominator. This enables us to cancel out "mol AgCl" and be left with "mol `~Cl_2` ".

Step 3: Convert moles of `~Cl_2` (from Step 2) to liters by multiplying by the conversion factor "1 mol = 22.4 L".

   3.991 mol `~Cl_2` x (22.4 L `~Cl_2` /1 mol `~Cl_2` ) = 89.40 L `~Cl_2`

Notice that the conversion factor "1 mol = 22.4 L" is oriented such that the "mole" part of the conversion factor is in the denominator. This enables us to cancel out moles and be left in liters.