The mass of [tex]YCl_3[/tex] that would be formed will be 18.22 grams
Stoichiometric calculations
Let us first look at the balanced equation of the reaction:
[tex]2Y + 3Cl_2 --- > 2YCl_3[/tex]
The mole ratio of Y to [tex]Cl_2[/tex] is 2:3.
Mole of 10.0 grams of Y = 10/88.9 = 0.11 moles
Mole of 10.0 grams [tex]Cl_2[/tex] = 10/71 = 0.14 moles
3/2 of 0.11 = 0.165. Thus, [tex]Cl_2[/tex] is limiting in availability.
Mole ratio of [tex]Cl_2[/tex] and [tex]YCl_3[/tex] = 3:2
Equivalent mole of [tex]YCl_3[/tex] = 2/3 x 0.14 = 0.093 moles.
Mass of 0.093 moles [tex]YCl_3[/tex] =0.093 x 195.26 = 18.22 grams
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