Question

What mass of YCL3 forms when 10.0 grams of each reactant are combined?

Answer 1

The mass of $YCl_3$ that would be formed will be 18.22 grams

Stoichiometric calculations

Let us first look at the balanced equation of the reaction:

$2Y + 3Cl_2 --- > 2YCl_3$

The mole ratio of Y to $Cl_2$ is 2:3.

Mole of 10.0 grams of Y = 10/88.9 = 0.11 moles

Mole of 10.0 grams  $Cl_2$ = 10/71 = 0.14 moles

3/2 of 0.11 = 0.165. Thus,  $Cl_2$ is limiting in availability.

Mole ratio of  $Cl_2$ and $YCl_3$ = 3:2

Equivalent mole of $YCl_3$ = 2/3 x 0.14 = 0.093 moles.

Mass of 0.093 moles $YCl_3$ =0.093 x 195.26 = 18.22 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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