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Based on the equation, how many grams of Br2 are required to react completely with 42.3 grams of AlCl3?

2AlCl3 + 3Br2 → 2AlBr3 + 3Cl2

66.5 grams
71.2 grams
76.1 grams
80.2 grams


Sagot :

Answer:

C.) 76.1 grams

Explanation:

To find the mass of bromine, you need to (1) convert grams AlCl₃ to moles AlCl₃ (via molar mass), then (2) convert moles AlCl₃ to moles Br₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Br₂ to grams Br₂ (via molar mass). It is important to arrange your ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).

Molar Mass (AlCl₃): 26.982 g/mol + 3(35.453 g/mol)

Molar Mass (AlCl): 133.332 g/mol

2 AlCl₃ + 3 Br₂ --> 2 AlBr₃ + 3 Cl₂

Molar Mass (Br₂): 2(79.904 g/mol)

Molar Mass (Br): 159.808 g/mol

42.3 g AlCl₃          1 mole               3 moles Br₂           159.808 g
------------------  x  -----------------  x  ----------------------  x  -------------------  =
                            133.332 g         2 moles AlCl₃            1 mole

=  76.0 g Br

*Our answers are slightly different most likely because we used slightly different molar masses*

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