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(05.04 MC)
Based on the equation, how many grams of Br₂ are
required to react completely with 42.3 grams of
AICI3? (5 points)
2AlCl3 + 3Br2 → 2AIBr3 + 3Cl₂
Select one:
O a.
66.5 grams
O b.
71.2 grams
O c. 76.1 grams
O d. 80.2 grams

Sagot :

Medunno13

Answer: C

Explanation:

  • The atomic mass of aluminum is 26.98154 g/mol.
  • The atomic mass of chlorine is 35.453 g/mol.

So, the formula mass of aluminum chloride is 26.98154+3(35.453)=133.34054 g/mol.

This means that 42.3 grams of aluminum chloride are 42.3/133.34054=0.31723285356426 moles.

From the equation, we know that for every 3 moles of bromine consumed, 2 moles of aluminum chloride are consumed.

This means that (0.31723285356426)(3/2)=0.47584928034639 moles of bromine are needed.

  • The atomic mass of one bromine atom is 79.904 g/mol

So, the formula mass of a bromine molecule is 2(79.904)=159.808 g/mol, and thus the answer is (159.808)(0.47584928034639), which is about 76.1 g

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