What volume of household bleach (6.0% NaOCl by mass) would be needed to react completely 200. mL of household ammonia which is 7.5% ammonia by mass?
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The oxygen from the bleach oxidises ammonia into nitrogen and water according to the equation:
3NaOCl + 2NH3 = 3NaCl + N2 + 3H2O
By the stoichiometry of this equation, 2×17 = 34 g ammonia reacts with 3×74.45 i.e. 223.5 g NaOCl.
So, 15 g ammonia should react with 223.35*15/34 = 98.54 g NaOCl.
The bleach solution is 6.0% NaOCl by mass. So, 6.0 g NaOCl is contained in 100 mL bleach solution (again, assuming the specific gravity of this solution too to be 1). Or, 98.54 g NaOCl should be obtained from 100*98.54/6 = 1642.3 mL bleach solution.
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