Question

For the following reaction, 5.10 grams of water are mixed with excess sodium. The reaction yields 8.17 grams of sodium hydroxide. sodium (s) + water (l) sodium hydroxide (aq) + hydrogen (g) What is the theoretical yield of sodium hydroxide ? grams What is the percent yield of sodium hydroxide ?

Answer 1

The balanced reaction will be

2Na(s) + 2H2O(l) ----------------- 2NaOH(aq) + H2(g)

Since sodium is present in excess, hence H2O is the limiting reagent

Number of moles of water = Mass/Molar mass = 5.10/18.01 = 0.28317 moles

2 moles of H2O will produce 2 moles of NaOH

Theoritical yield of NaOH in moles = 0.28317 moles

Molar mass of NaOH = Molar mass of Na + Molar mass of O + Molar mass of H = 23 + 16 + 1 = 40 g/mol

Theoritical yield of NaOH = number of moles * Molar mass = 0.28317 mol * 40 g/mol = 11.33 grams

Percent Yield = Actual Yield/Theoritical Yield * 100 = 8.17/11.33 * 100 = 72.1%

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