Question

In: Chemistry

If 25.0 mL of a 0.238M FeO4^2- solution is allowed to react with 55.0 mL of...

If 25.0 mL of a 0.238M FeO4^2- solution is allowed to react with 55.0 mL of 1.47 M aqueous ammonia, what volume of nitrogen gas can form at 25°C and 1.50 atm?

Volume = mL

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Expert Solution

Ans. Step 1: Determine the limiting reactant, if any:

# Balanced reaction:

2 FeO42-(aq) + 2 NH3(aq) + 10 H+(aq) ---> N2(g) + 2 Fe3+(aq) + 8 H2O(l)

Theoretical molar ratio of reactants = FeO42- : NH3 = 2 : 2 = 1 : 1

# Moles of FeO42- = Molarity x Vol. in liters = 0.238 M x 0.025 L = 0.00595 mol

Moles of NH3 = 1.47 M x 0.055 L = 0.08085 mol

Experimental molar ratio of reactants = FeO42- : NH3 = 0.00595 : 0.08085 = 0.07 : 1

# Comparing the theoretical and experimental molar ratios of reactants, the experimental moles of FeO42- is less than its theoretical value of 1 mol while keeping that of NH3 constant at 1 mol. So, FeO42- is the limiting reactant.

# Step 2: Determine N2 yield: The formation of product follows the stoichiometry of the limiting reactant.

Following stoichiometry of balanced reaction, 2 mol FeO42- forms 1 mol N2.

So, moles of N2 formed = (½) x 0.00595 mol = 0.002975 mol

# Step 3: Calculate the volume of N2 gas using ideal gas equation –


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