How many molecules of chlorine gas would be produced if 10.0g of aluminum chloride decomposed into aluminum precipitate and chlorine gas?

Question

Olivia Anton · Accepted Answer

6.77 x #10^22# molecules of #Cl_2#.

First you need to write the correct BALANCED chemical reaction equation. Then you can determine how many moles of #Cl_2# would be produced from the available moles of Cl in 10.0g of #AlCl_3#. Finally, you will convert this value into the number of molecules using Avogadro's number. #2AlCl_3 → 2Al + 3Cl_2# So, TWO moles of #AlCl_3# will produce THREE moles of #Cl_2#. The molecular weight of #AlCl_3# is 133.33g/mol, so 10.0g is 10.0/133.33 = 0.075mol With our equation ratio of 3/2 this will produce 0.1125 mol #Cl_2#. 0.1125 * 6.022 x #10^23# molecules/mole = 6.77 x #10^22# molecules of #Cl_2#.

Jose Ayala · Answer

undefined To determine the number of molecules of chlorine gas produced, first, calculate the moles of aluminum chloride using its molar mass. Then, since aluminum chloride decomposes to form two moles of chlorine gas for every mole of aluminum chloride, use stoichiometry to find the moles of chlorine gas produced. Finally, convert moles of chlorine gas to molecules using Avogadro's number.

1. Calculate moles of aluminum chloride:
   $ \text{Molar mass of } AlCl_3 = 26.98 + 3 \times 35.45 = 133.33 \, \text{g/mol} $
   $ \text{Moles of } AlCl_3 = \frac{10.0 \, \text{g}}{133.33 \, \text{g/mol}} $

2. Use stoichiometry to find moles of chlorine gas:
   $ \text{From the balanced equation:} \, 2 \, \text{moles of } AlCl_3 \rightarrow 3 \, \text{moles of } Cl_2 $
   $ \text{Moles of } Cl_2 = \frac{\text{Moles of } AlCl_3}{2} \times 3 $

3. Convert moles of chlorine gas to molecules:
   $ \text{Number of molecules} = \text{Moles of } Cl_2 \times 6.022 \times 10^{23} $

Perform the calculations to find the number of molecules of chlorine gas produced.