Question

How many grams of 0.291 wt% aqueous NaOH (FM = 40.00) are required to provide a 30% excess to react with 75.0 mL of 0.0413 M Al3+ by the reaction Al3+ + 3OH- → Al(OH)3 (s) ?

Answer 1

Al3+ + 30147 - 9.) 09.) Al (OH), s moles of cn = 3 *moles of Al3t 234 0.075X 0.0413 mol = 0.0093 mol 30% excess - 0.60934 1.3 o.0121 mol Let w e why of soln = (0.291xW) - 0.0121 40 =) WS 0.012) X Yox 100 0.291 = 166 g.