Household bleach contains 5.0% sodium hypochlorite (NaOCl,74.44 g/mol)by mass.Household ammonia contains 5.0% NH₃ (17.04 g/mol)by mass.If they are mixed,toxic chloramine gas (NH₂Cl; 51.48 g/mol)is produced.What volume of NH₂Cl gas results when 50.00 g of bleach and 50.00 g of 5.0% ammonia are combined? Assume 1 mol NH₂Cl occupies 24.4 L. NaOCl(aq)+ NH₃(aq)$\to$ NH₂Cl(aq)+ NaOH(aq)

Question

Quizplus · Accepted Answer

1. Convert mass of NaOCl and NH3 to moles: - Moles of NaOCl = 50.00 g / 74.44 g/mol = 0.672 mol - Moles of NH3 = (5.0/100) x 50.00 g / 17.04 g/mol = 0.147 mol 2. Determine the limiting reagent: - Based on the equation, the stoichiometric ratio between NaOCl and NH3 is 1:1 (1 mol of NaOCl reacts with 1 mol of NH3) - Therefore, NH3 is the limiting reagent because there is less of it than NaOCl 3. Calculate the moles of NH₂Cl produced: - From the equation, 1 mol of NH₂Cl is produced for every 1 mol of NH3 used - Moles of NH₂Cl = 0.147 mol 4. Convert moles of NH₂Cl to volume: - 1 mol of gas occupies 24.4 L - Volume of NH₂Cl produced = 0.147 mol x 24.4 L/mol = 3.59 L = 0.82 L (rounded to 2 significant figures) Therefore, the answer is C.

Household Bleach Contains 5 →\to→ NH2Cl(aq)+ NaOH(aq) A)2

Household Bleach Contains 5 $\to$ NH₂Cl(aq)+ NaOH(aq)
A)2