Answer to Question #325758 in Chemistry for Lynn

Question

TEST 2. PROBLEM SOLVING

Solve the problem below. Show your answers legibly, concisely, and completely. Use the back portion if necessary.

1. Determine the limiting reactant and excess reactant, when 36g of Al and 28.5g of HBr are reacted and how many grams of H_{2} gas are formed?

Al + HBr -> AlB*r_{3} + H_{2}

Given:

Find:

Solution:

Answer 1

Given:

Mass of Al = 36 g

Mass of HBr = 28.5 g

Find:

The limiting reactant is ???

The excess reactant is ???

Mass of H₂ = ???

Solution:

The molar mass of Al is 26.98 g/mol

The molar mass of HBr is 80.91 g/mol

Calculate the moles of each reactant:

Moles of Al = (36 g Al) × (1 mol Al / 26.98 g Al) = 1.334 mol Al

Moles of HBr = (28.5 g HBr) × (1 mol HBr / 80.91 g HBr) = 0.352 mol HBr

Balanced chemical equation:

2Al + 6HBr → 2AlBr₃ + 3H₂

According to stoichiometry:

2 mol of Al react with 6 mol of HBr

Thus, 1.334 mol of Al react with:

(1.334 mol Al) × (6 mol HBr / 2 mol Al) = 4.002 mol HBr

However, initially there is 0.352 mol of HBr (according to the task)

Therefore, HBr acts as limiting reactant and Al is excess reactant

According to stoichiometry:

6 mol of HBr produce 3 mol H₂

Thus, 0.352 mol HBr produce:

(0.352 mol HBr) × (3 mol H₂ / 6 mol HBr) = 0.176 mol H₂

The molar mass of H₂ is 2.016 g/mol

Therefore,

Mass of H₂ = (0.176 mol H₂) × (2.016 g H₂ / 1 mol H₂) = 0.355 g

Mass of H₂ = 0.355 g

Answer:

The limiting reactant is HBr

The excess reactant is Al

0.355 grams of H₂ gas are formed