Question

Ammonia ( NH3, MM 17.031 g/mol) and hypobromite ( OBr−) react according to the following chemical reaction. 2NH3+3OBr−⟶N2+3Br−+3H2O Suppose 0.325 g of ammonia reacts with 14.5 mL of a hypobromite solution. Calculate the concentration of the hypobromite solution.

Answer 1

Answer:

1.98 M

Explanation:

Step 1: Write the balanced reaction

2 NH₃ + 3 BrO⁻⟶ N₂ + 3 Br⁻ + 3 H₂O

Step 2: Calculate the moles corresponding to 0.325 g of ammonia

The molar mass of NH₃ is 17.03 g/mol.

0.325 g × (1 mol/17.03 g) = 0.0191 mol

Step 3: Calculate the reacting moles of hypobromite

The molar ratio of NH₃ to BrO⁻ is 2:3. The reacting moles of hypobromite are 3/2 × 0.0191 mol = 0.0287 mol

Step 3: Calculate the molar concentration of the hypobromite solution

M = 0.0287 mol / 0.0145 L = 1.98 M

Answer 2

The concentration of the hypobromite solution (BrO⁻) needed for the reaction is 1.97 M

We'll begin by calculating the number of mole in 0.325 g of NH₃. This can be obtained as follow:

Mass of NH₃ = 0.325 g

Molar mass of NH₃ = 17.031 g/mol

Mole of NH₃ =?

Mole = mass / molar mass

Mole of NH₃ = 0.325 / 17.031

Mole of NH₃ = 0.01908 mole

Next, we shall determine the number of mole of BrO⁻ needed to react with 0.01908 mole of NH₃. This can be obtained as follow:

2NH₃ + 3BrO⁻ —> N₂ + 3Br⁻ + 3H₂O

From the balanced equation above,

2 moles of NH₃ reacted with 3 moles of BrO⁻.

Therefore,

0.01908 mole of NH₃ will react with = $\frac{0.01908 * 3 }{2}$ = 0.02862 mole of BrO⁻.

Finally, we shall determine the concentration of the BrO⁻ solution.

Mole of BrO⁻ = 0.02862 mole

Volume of BrO⁻ solution = 14.5 mL = 14.5 / 1000 = 0.0145 L

Concentration of BrO⁻ =?

Concentration = mole / Volume

Concentration of BrO⁻ = 0.02862 / 0.0145

Concentration of BrO⁻ = 1.97 M

Therefore, the concentration of the hypobromite solution (BrO⁻) is 1.97 M

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