What mass of aluminum chloride is required to produce 0.125

Last updated: 7/31/2022

What mass of aluminum chloride is required to produce 0.125 moles of aluminum sulfide according to the following equation? AlCl3 = 133.34 g/mol 2AICI3(aq) + 3H₂S(aq) → Al₂S3(s) + 6HCl(aq) 16.7 g 66.6 g 18.8 g 33.3 g